["14.16 Chapter 14 a(x - x1) + b(y - y1) = 0 \u21d2 4(x + 3) - 3(y - 2) = 0 \u21d2 4x - 3y + 18 = 0. Hence, the equation of the required line is 4x - 3y + 18 = 0. Example 14.24 Find the equation of a line passing through point (-2, 3) and perpendicular to 7x + 2y + 3 = 0. Solution Here, (x1, y1) = (-2, 3), a = 7 and b = 2. \u2234 Equation of the line perpendicular to 7x + 2y + 3 = 0 and passing through (-2, 3) is b(x - x1) - a(y - y1) = 0. That is, 2(x + 2) -7(y - 3) = 0 \u21d2 2x - 7y + 25 = 0. Hence, the required equation of the line is 2x - 7y + 25 = 0. Example 14.25 The line (8x + 3y - 15) + l(3x - 8y + 2) = 0 is parallel to X-axis. Find l. Solution The given line is (8x + 3y - 15) + l(3x -8y + 2) = 0. That is, x(8 + 3l) + y(3 - 8l) + (2l - 15) = 0. Since the given line is parallel to X-axis, its slope = 0. \u2212(8 + 3\u03bb ) = 0 3 \u2212 8\u03bb \u21d2 8 + 3l = 0 \u22128 Hence, l = 3 . Example 14.26 The equation of the line passing through the point of intersection of lines 2x \u2212 y + 3 = 0 and 3x + y + 7 = 0 and perpendicular to 2x \u2212 3y + 4 = 0, is ________. (a) 3x + 2y \u2212 7 = 0\u2003\u2003 (b) 3x + 2y + 8 = 0\u2003\u2003 (c) 3x + 2y \u2212 8 = 0\u2003\u2003 (d) 3x \u2212 2y + 1 = 0 Hints (i)\t Find m and the intersection point. Then use slope-point form. (ii)\t Find the common point (x1, y1) of first two equations. (iii)\t Find the slope (m) of third line. through (x1, y1) and having slope \uf8eb \u2212 1\uf8f6 . (iv)\t Find the equation of the line passing \uf8ec\uf8ed m \uf8f7\uf8f8","Coordinate Geometry 14.17 Example 14.27 The area of the figure formed by |x| + |y| = 2 is______. (in sq. units) (a) 2\u2003\u2003 (b) 4\u2003\u2003(c) 6\u2003\u2003 (d) 8 HINTS (i)\t Plot the figure. (ii)\t Find the intercepts made by given line. of the triangle is ab . (iii)\t If the intercepts are a and b, then the area 2 Example 14.28 The sum of the reciprocals of the intercepts of a line is 1 , then the line passes through the 2 point is______. (a) (1, 1)\u2003\u2003 (b) (2, 1)\u2003\u2003 (c) \uf8eb 1 , 1 \uf8f6 \u2003\u2003(d) (2, 2) \uf8ec\uf8ed 4 4 \uf8f8\uf8f7 HINTS (i)\t Use x + y = 1. a b (ii)\t Solve, 111 and get the relation between a and b. a+b=2 (iii)\t Use the formula x + y = 1. a b Mid-point Let A(x1, y1) and B(x2, y2) be two given points and M be the mid-point of AB. Then, M = \uf8eb x1 + x2 , y1 + y2 \uf8f6 . \uf8ed\uf8ec 2 2 \uf8f8\uf8f7 Hence, the coordinates of the mid-point of the line segment joining the points (x1, y1) and (x2, y2) are given by \uf8eb x1 + x2 , y1 + y2 \uf8f6 . \uf8ed\uf8ec 2 2 \uf8f7\uf8f8 Example 14.29 Find the mid-point of the line segment joining the points (2, -6) and (6, -4). Solution Let A(2, -6) and B(6, -4) be the given points and M be the mid-point of AB. Then, M = \uf8eb x1 + x2 , y1 + y2 \uf8f6 \uf8ec\uf8ed 2 2 \uf8f7\uf8f8 = \uf8eb 2 + 6 , \u22126 + (\u22124)\uf8f6 = (4, \u22125) . \uf8ed\uf8ec 2 2 \uf8f8\uf8f7 Hence, the mid-point of AB is (4, \u22125).","14.18 Chapter 14 Centroid Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of DABC, and G be its centroid. Then, the coordinates of G are given by, G = \uf8eb x1 + x2 + x3 , y1 + y2 + y3 \uf8f6 . \uf8ec\uf8ed 2 2 \uf8f7\uf8f8 Example 14.30 Find the centroid of DABC whose vertices are A(2, -3), B(4, 2) and C(-3, -2). Solution Given, A(2, -3), B(4, 2) and C(-3, -2). So, centroid of DABC \uf8eb x1 + x2 + x3 , y1 + y2 + y3 \uf8f6 = \uf8eb 2 + 4 \u2212 3 , \u22123 +2 \u2212 2 \uf8f6 = (1, \u22121). \uf8ec\uf8ed 2 2 \uf8f8\uf8f7 \uf8ec\uf8ed 3 3 \uf8f8\uf8f7 Hence, (1, -1) is the centroid of DABC. Example 14.31 Find the third vertex of DABC, if two of its vertices are A(-2, 3), B(4, 5) and its centroid is G(1, 2). Solution Let C(x, y) be the third vertex. Given, centroid of DABC = (1, 2) \u21d2 \uf8eb \u22122 +4 + x, 3+5+ y \uf8f6 = (1, 2) \uf8ec\uf8ed 3 3 \uf8f8\uf8f7 \u21d2 \uf8eb x + 2 , y + 8 \uf8f6 = (1, 2) \uf8ec\uf8ed 3 3 \uf8f8\uf8f7 \u21d2 x + 2 = 1, y +8 =2 3 3 \u21d2 x =1, y = \u22122. \u2234 The third vertex is (1, -2). \u2002Notes\u2002 1.\t I\u0007f the mid-points of the sides BC, AC and AB of DABC, respectively, are P(x1, y1), Q(x2, y2) and R(x3, y3), then its vertices are A(\u2212x1 + x2 + x3, \u2212y1 + y2 + y3), B(x1 \u2212 x2 + x3, y1 \u2212 y2 + y3) and C(x1 + x2 \u2212 x3, y1 + y2 \u2212 y3). 2.\t \u0007The fourth vertex of a parallelogram whose three consecutive vertices are (x1, y1), (x2, y2) and (x3, y3) when taken in order is (x1 \u2212 x2 + x3, y1 \u2212 y2 + y3).","Coordinate Geometry 14.19 Example 14.32 Find the fourth vertex of the parallelogram whose three consecutive vertices are (8, 8), (6, 1) and (-1, 1). Solution Let the three vertices of the parallelogram be A(8, 8), B(6, 1) and C(-1, 1), then fourth vertex D(x, y) is given by D(x, y) = (x1 - x2 + x3, y1 - y2 + y3) = (8 - 6 - 1, 8 - 1 + 1) = (1, 8). Hence, the fourth vertex is D(1, 8). Example 14.33 If the centroid of a triangle is (6, 6) and its ortho-centre is (0, 0), then find its circum-centre. (a) (3, 3)\u2003\u2003 (b) (6, 6)\u2003\u2003 (c) (9, 9)\u2003\u2003 (d) (12, 12) Solution Ortho-centre, centroid and circum-centre are collinear. We know that centroid divides the line segment joining the ortho-centre, centroid and circum- centre (OGS ) in the ratio 2 : 1 from the ortho-centre (O). Let S(x, y), G(6, 6) and O(0, 0) (6, 6) = \uf8eb 2x \u00d7 1\u00d7 0 , 2 \u00d7 y\u00d7 0 \uf8f6 \uf8ed\uf8ec 2+1 2 +1 \uf8f7\uf8f8 (6, 6) = \uf8eb 2x , 2y \uf8f6 \uf8ed\uf8ec 3 3 \uf8f8\uf8f7 \u21d2 2x = 6, 2y =6 3 3 \u21d2 x = 6 and y = 9. \u2234 The circum-centre is (9, 9). Example 14.34 C(3, 0) and D(3, 1) are the points of trisection of a line segment AB. Find the respective coordinates of A and B. (a) (3, 2), (3, 0)\u2003\u2003 (b) (3, \u22121), (3, 2)\u2003\u2003 (c) (\u22123, 1), (3, 2)\u2003\u2003 (d) None of these Solution Let A and B be (a1, b1) and (a2, b2). Given, C(3, 0) and D(3, 1) are the points of trisection of AB. AB CD","14.20 Chapter 14 \u21d2 C is the mid-points of AD and D is the mid-points of CB. \u21d2 (3, 0) = \uf8eb a1 + 3 , b1 + 1 \uf8f6 \uf8ed\uf8ec 2 2 \uf8f8\uf8f7 \u21d2 a1 = 3 and b2 = \u22121. Also, (3, 1) = \uf8eb 3 + a2 , 0 + b2 \uf8f6 \uf8ec\uf8ed 2 2 \uf8f8\uf8f7 \u21d2 a2 = 3 and b2 = 2. \u2234 The coordinates of A and B are (3, \u22121) and (3, 2).","Coordinate Geometry 14.21 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t If x > 0 and y < 0, then the point (x, \u2212y) lies in \t16.\t The area of triangle formed by the line y = mx + c ______ quadrant. with the coordinate axes is ______. \t2.\t Which point among (2, 3), (\u22123, \u22124) and (1, \u22127) is 1\t 7.\t The points (2, 3), (\u22121, 5) and (x, \u22122) form a straight nearest to the origin? line, then x is _____. \t3.\t The lines 2y + 3 = 0 and x = 3 intersect at ______. \t18.\t If the point (x, y) lies in the second quadrant, then x is ______ and y is ______. \t4.\t The points (0, 0), (0, 4) and (4, 0) form a\/an ______ triangle. \t19.\t The angle between lines x = 5 and x = 7 is ______. \t5.\t A linear equation in two variables is always a ______. \t20.\t The point of intersection of X-axis and 3x + 2y \u22125 = 0 is ______. \t6.\t The slope of the line ax + by + c = 0 is _______. \t21.\t If a = 0, then the line ax + by + c = 0 is parallel to \t7.\t If (x, y) represents a point and |x| > 0 and y < 0, ______. then in which quadrant(s) can the point lie? 2\t 2.\t The lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 \t8.\t The equation of a line parallel to Y-axis and passing are perpendicular to each other, then _______. through (\u22123, \u22124) is ______. 2\t 3.\t The point of intersection of X-axis and Y-axis is \t9.\t The slope of line perpendicular to the line joining _______. points (2, 3) and (\u22122, 5) is ______. \t24.\t The line y = k is parallel to _____ axis. \t10.\t The slope of altitude from A to BC of triangle A(2, 3), B(-3, 2) and C(3, 5) is ______. \t25.\t A, B and C are three points such that AB = AC + CB, then A, B and C are _______. \t11.\t If the line x + y = m passes through origin, then a b 2\t 6.\t The line ax + by + c = 0 meets Y-axis at _______ point. PRACTICE QUESTIONS the value of m is _______. 2\t 7.\t If slope of a line (l) is tan\u2009q, then slope of a line 1\t 2.\t If (x, y) represents a point and xy > 0, then the perpendicular to (l) is ______. point may lie in _______ or ______quadrant. \t28.\t The lines x = 2 and y = \u22123 intersect in ______ 1\t 3.\t The slope-intercept form of the line 2x + 3y + 5 quadrant. = 0 is ______. \t29.\t The slope of a line which is parallel to the line \t14.\t The lines 3x + 2y + 7 = 0 and 6x + 4y + 9 = 0 are making an inclination of 45\u00ba with positive X-axis ________to each other. is _____. 1\t 5.\t The points (p, q + r), (q, r + p) and (r, q + p) are 3\t 0.\t If the slope of two lines are equal, then the lines are ______. ______. Short Answer Type Questions \t31.\t Find the equation of a line passing through points 3\t 4.\t Let A(\u22123, 2), B(4, 1) and C(\u22122, k) be three points A(\u22122, 3) and B(4, 7). such that AC = BC. Find the value of k. \t32.\t Find the area of the circle passing through (\u22122, 3) 3\t 5.\t Find the distance between points (2, \u22123) and with centre (5, 2). (4, 6). \t33.\t If (2x + 3y + 1) + l(x \u2212 2y \u2212 3) = 0 represents the \t36.\t If the line 2x \u2212 ky + 6 = 0 passes through the point equation of a horizontal line, then find the value of l. (2, \u22128), then find the value of k.","14.22 Chapter 14 \t37.\t Find the area of square, whose diagonally opposite 4\t 2.\t Find the equation of a line having inclination 60\u00b0 vertices are (\u22122, 3) and (4, 5). and making an intercept of \u22121 on Y-axis. 3\t 8.\t If A(a + b, a \u2212 b) and B(\u2212a + b, \u2212a \u2212 b), then find 3 the distance AB. 4\t 3.\t Find the point on X-axis, which is equidistant \t39.\t Find the intercepts made by the line 3x \u2212 2y \u2212 6 = from A(6, 3) and B(\u22121, 4). 0 on the coordinate axes. \t44.\t Show that the points (\u22121, \u22121), (6, 1), (8, 8) and \t40.\t Find the inclination of the line\u2009 3x \u2212 3y + 6 = 0. (1, 6), when joined in the given order form a rhombus. \t41.\t Find the circum-centre of the triangle whose ver- tices are A(\u22123, \u22121), B(1, 2) and C(0, \u22124). \t45.\t Find the equation of a line, whose y-intercept is \u22125 and passes through point A(\u22123, 2). Essay Type Questions \t46.\t Find the equations of the lines whose intercepts \t49.\t Find the equations of a line which forms area 5 sq. are the roots of the equation 4x2 \u2212 3x \u2212 1 = 0. units with the coordinate axes and having sum of intercepts is 7. 4\t 7.\t The equation of one of the diagonals of a rhombus is 3x + 4y \u2212 7 = 0. Find the equation of the other 5\t 0.\t If points A(1, 6), B(5, 2) and C(12, 9) are three diagonal passing through (-1, \u22122). consecutive vertices of a parallelogram, then find the equation of the diagonal BD. \t48.\t Find the equation of the line passing through (\u22125, 11) and making equal intercepts, but opposite in magnitude on the coordinate axes. PRACTICE QUESTIONS CONCEPT APPLICATION Level 1 \t1.\t If (1, -3), (-2, -3) and (-2, 2) are the three verti- \t\t(a) y = \u22125\t\t (b) x = 7 ces of a parallelogram taken in that order, then the fourth vertex is ______. \t\t(c) x = \u22125\t\t (d) y = 7 \t\t(a) (-1, -2)\t\t\t(b)(1, 2) \t5.\t If (2, 0) and (\u22122, 0) are the two vertices of an equi- lateral triangle, then the third vertex can be ______. \t\t(c) (-1, 2)\t\t\t (d) (1, -2) \t\t(a) (0, 0)\t\t (b) (2, \u22122) \t2.\t Find the equation of the line that passes through \t\t(c) (0,2 3 ) \t(d) ( 3, 3 ) point (5, -3) and makes an intercept 4 on the X-axis. \t6.\t The points (a, b + c), (b, c + a) and (c, a + b) \t\t(a) 3x - y + 12 = 0\t\t (b) 3x + y + 12 = 0 \t\t(a) are collinear. \t\t(b) form a scalene triangle. \t\t(c) 3x - y - 12 = 0\t\t (d) 3x + y - 12 = 0 \t\t(c) form an equilateral triangle. \t\t(d) None of the above. \t3.\t The inclination of line x - 3y + 1 = 0 with the positive X-axis is ______. \t\t(a) 60\u00b0\t\t\t(b)30\u00b0 \t7.\t The equation of the line making equal intercepts and passing through the point (-1, 4) is ______. \t\t(c) 45\u00b0\t\t\t (d) 90\u00b0 \t4.\t The equation of the line perpendicular to Y-axis \t\t(a) x - y = 3\t (b) x + y + 3 = 0 and passing through point (\u22125, 7) is ______. \t\t(c) x + y = 3\t (d) x - y + 3 = 0","Coordinate Geometry 14.23 \t8.\t The endpoints of the longest chord of a circle are \t16.\t The two lines 3x + 4y - 6 = 0 and 6x + ky - 7 = 0 (-4, 2) and (-6, -8). Find its centre. are such that any line which is perpendicular to the first line is also perpendicular to the second line.\t \t\t(a) \uf8eb \u2212 10 , \u22122 \uf8f6 \t\t (b)(-5, -2) Then, k = ___. \uf8ed\uf8ec 3 \uf8f8\uf8f7 \t\t(a) -8\t\t (b) -6 \t\t(c) (-5, -4)\t\t\t(d) (-5, -3) \t\t(c) 6\t\t (d) 8 \t9.\t The equation of the line passing through point (\u22123, \u22127) and making an intercept of 10 units on 1\t 7.\t The line x = my, where m < 0, lies in the quadrants. X-axis can be _____. \t\t(a) 1st, 2nd \t\t (b) 2nd, 4th \t\t(a) 4x + 3y = \u22129\t\t (b) 8x \u2212 3y = 80 \t\t(c) 3rd, 4th\t\t (d) 3rd, 1st \t\t(c) 7x \u2212 13y \u2212 70 = 0\t (d) 7x + 3y \u2212 70 = 0 1\t 8.\t Find the area in square units, of the rhombus with vertices (2, 1), (-5, 2), (-4, -5) and (3, -6), taken 1\t 0.\t The points on the Y-axis which are at a distance of in that order. 5 units from (4, \u22121) are _____. \t\t(a) (0, \u22122), (0, 4)\t\t (b) (0, 2), (0, -4) \t\t(a) 24\t\t (b) 48 \t\t(c) (0, 2) (0, 4)\t\t (d) (0, \u22122) (0, \u22124) \t\t(c) 36\t\t (d) 50 \t11.\t If the slope and the y-intercept of a line are the 1\t 9.\t The radius of a circle with centre (-2, 3) is 5 units, roots of the equation x2 \u2212 7x \u2212 18 = 0, then the then the point (2, 5) lies _____. equation of the line can be _____. \t\t(a) on the circle \t\t(a) 2x + y \u2212 9 = 0\t\t (b)2x \u2212 y + 9 = 0 \t\t(b) inside the circle \t\t(c) outside the circle \t\t(c) 9x + y + 2 = 0\t\t (d) 9x + 2y \u2212 2 = 0 \t\t(d) None of the above 1\t 2.\t If the points (k, k \u2212 1), (k + 2, k + 1) and (k, k + 3) \t20.\t One end of the diameter of a circle with the cen- are three consecutive vertices of a square, then its tre as origin is (-2, 10). Find the other end of the area (in square units) is _____. diameter. \t\t(a) 2\t\t\t (b) 4 PRACTICE QUESTIONS \t\t(c) 8\t\t\t (d) 6 \t\t(a) (-2, -10)\t (b)(0, 0) 1\t 3.\t The equation of the line making intercepts of \t\t(c) (2, -10)\t\t (d) (2, 10) equal magnitude and opposite signs, and passing through the point (\u22123, \u22125) is ______. \t21.\t If the roots of the quadratic equation x2 - 7x + 12 = 0 are intercepts of a line, then the equation of the \t\t(a) x \u2212 y = 2\t\t (b) 2x + y = \u22124 line can be _____. \t\t(c) 3x + 3y = 6\t\t (d) x \u2212 y = \u221210 \t\t(a) 2x + 3y = 6\t (b) 4x + 3y = 12 \t14.\t If the endpoints of the diameter of a circle are (-2, \t\t(c) 4x + 3y = 6\t (d) 3x + 4y = 6 3) and (6, -3), then the area of the circle (in square units) is _____. 2\t 2.\t Find the value of l, if the line x \u2212 3y + 4 + l(8x \u2212 3y + 2) = 0 is parallel to the X-axis. 550 540 \t\t(a) 3 \t\t\t(b) 7 \t\t(a) 1 \t\t (b) 5 5 8 560 550 \t\t(c) 7 \t\t\t(d) 7 \t\t(c) 3 \t\t (d) 1 8 8 \u2212 \u2212 \t15.\t The inclination of the line 3x - y + 3 = 0 with \t23.\t The slope of the line joining the points (2, k - 3) the positive X-axis is ______. and (4, -7) is 3. Find k. \t\t(a) 30\u00b0\t\t\t(b)45\u00b0 \t\t(a) -10\t\t (b) -6 \t\t(c) 60\u00b0\t\t\t (d) 90\u00b0 \t\t(c) -2\t\t (d) 10","14.24 Chapter 14 \t24.\t The angle between the lines x = 10 and y = 10 is \t\t(a) 1st quadrant\t (b) 2nd quadrant ______. \t\t(c) 3rd quadrant\t (d) 4th quadrant \t\t(a) 0\u00b0\t\t (b) 90\u00b0 2\t 8.\t The circum-centre of the triangle formed by points O(0, 0), A(6, 0) and B(0, 6) is _____. \t\t(c) 180\u00b0\t\t (d) None of these \t25.\t The two lines 5x + 3y + 7 = 0 and kx \u2212 4y + 3 = 0 \t\t(a) (3, 3)\t\t (b) (2, 2) are perpendicular to the same line. Find the value of k. \t\t(c) (1, 1)\t\t (d) (0, 0) \t\t(a) \u2212 20 \t\t (b) \u2212 20 2\t 9.\t The lines 3x \u2212 y + 2 = 0 and x + 3y + 4 = 0 inter- 7 3 sect each other in the _____. 20 12 \t\t(a) 1st quadrant\t (b) 4th quadrant 9 5 \t\t(c) \t\t (d) \t\t(c) 3rd quadrant\t (d) 2nd quadrant \t26.\t The lines x \u2212 2y + 3 = 0, 3x \u2212 y = 1 and kx \u2212 y \t30.\t Centre of the circle is (a, b). If (0, 3) and (2, 0) + 1 = 0 are concurrent. Find k. are two points on a circle, then find the relation 1 between a and b. \t\t(a) 1\t\t (b) 2 \t\t(a) 4a - 6b - 5 = 0 3 5 \t\t(c) 2 \t\t (d) 2 \t\t(b) 4a + 6b - 5 = 0 \t27.\t Find the quadrant in which the lines 2x + 3y - 1 = \t\t(c) -4a + 5 = 0 0 and 3x + y - 5 = 0 intersect each other. \t\t(d) 4a - 6b + 5 = 0 Level 2 \t31.\t The equation of a line passing through P(3, 4), such \t\t(a) 4x + 7y \u2212 20 = 0 that P bisects the part of it intercepted between the \t\t(b) 3x \u2212 7y + 3 = 0 coordinate axes is _____. \t\t(c) 3x + 2y + 15 = 0 \t\t(d) 3x \u2212 7y \u2212 15 = 0 PRACTICE QUESTIONS \t\t(a) 3x + 4y = 25\t (b) 4x + 3y = 24 \t\t(c) x \u2212 y = -1\t(d) x + y = 7 \t35.\t Find the area of the triangle formed by the line 3x - 4y + 12 = 0 with the coordinate axes. 3\t 2.\t The line 7x + 4y = 28 cuts the coordinate axes at A and B. If O is the origin, then the ortho-centre \t\t(a) 6 units2\t\t (b) 12 units2 of DOAB is _____. \t\t(c) 1 units2\t\t (d) 36 units2 \t\t(a) (4, 0)\t\t (b) (0, 7) \t36.\t The line joining the points (2m + 2, 2m) and (2m + 1, 3) passes through (m + 1, 1), if the values of m \t\t(c) (0, 0)\t\t (d) None of these are _____. 3\t 3.\t If the roots of the quadratic equation x2 \u2212 5x + 6 = \t\t(a) 5, \u2212 1 \t\t (b) 1, \u22121 0 are the intercepts of a line, then the equation of 5 the line can be _____. \t\t(a) 2x + 3y = 6 \t\t(c) 2, \u2212 1 \t\t (d) 3, \u2212 1 \t\t(b) 3x + 2y = 6 2 3 \t\t(c) Either (a) or (b) \t\t(d) None of these \t37.\t The length (in units) of the line joining the points (4, 3) and (\u22124, 9) intercepted between the coordi- \t34.\t The equation of the line whose x-intercept is 5, nate axes is ______. and which is parallel to the line joining the points (3, 2) and (\u22124, \u22121) is _____. \t\t(a) 10\t\t (b) 8 \t\t(c) 6\t\t (d) 4","Coordinate Geometry 14.25 \t38.\t The equation of a line parallel to 8x \u2212 3y + 15 = 0 \t\t(D) L\u0007 et the fourth vertex be D(x, y). We know that and passing through the point (\u22121, 4) is _____. the diagonals of a parallelogram bisect each other. \t\t(a) 8x \u2212 3y \u2212 4 = 0 \t\t(a) ADCB\t \t (b) DCAB \t\t(b) 8x \u2212 3y \u2212 20 = 0 \t\t(c) 8x \u2212 3y + 4 = 0 \t\t(c) DACB\t\t (d) CDAB \t\t(d) 8x \u2212 3y + 20 = 0 4\t 4.\t If A = (1, \u22126), B = (5, \u22122) and C = (12, \u22129) are the \t39.\t (0, 0),(3, 3 ) and (0, 2 3 ) are the three vertices three consecutive vertices of a parallelogram, then of a triangle. The distance between the ortho- find the fourth vertex. The following are the steps centre and the cirum-centre of the triangle is involved in solving the above problem. Arrange _______. (in units) them in sequential order from beginning to end. \t\t(A) \u00075 + x = 13 , \u22122 + y = \u221215 \u21d2 x = 8, and y = 2 2 2 2 \t\t(a) 3 \t\t (b) 5 -13. Therefore, D = (8, \u221213). \t\t(c) 6 \t\t (d) 0 \t\t(B) \u2234 \uf8eb 5+ x , \u22122 + y\uf8f6 = \uf8eb 1+ 12 , \u22126 \u2212 9\uf8f6 . \uf8ec\uf8ed 2 2 \uf8f7\uf8f8 \uf8ed\uf8ec 2 2 \uf8f8\uf8f7 4\t 0.\t In a parallelogram PQRS, P(15, 9), Q(7, 10), R(\u22125, \u22124), then the fourth vertex S is _______. \t\t(C) Let the fourth vertex be D = (x, y). \t\t(a) (3, \u22122)\t\t (b) (3, \u22124) \t\t(D) W\u0007 e know that diagonals of a parallelogram bisect each other. \t\t(c) (9, \u22125)\t\t (d) (3, \u22125) 4\t 1.\t If the roots of the quadratic equation 3x2 \u2212 2x \u2212 1 \t\t(a) ACBD\t \t (b) ABDC = 0 are the intercepts of a line, then the line can be ______. \t\t(c) CBDA\t \t (d) CDBA \t\t(a) x \u2212 3y \u2212 1 = 0 \t45.\t Find the product of intercepts made by the line 7x \t\t(b) 3x \u2212 y + 1 = 0 - 2y - 14 = 0 with coordinate axes. \t\t(c) Either (a) or (b) \t\t(d) None of these \t\t(a) -7\t\t (b) 2 \t\t(c) 14\t\t (d) -14 4\t 2.\t The length (in units) of a line segment intercepted 4\t 6.\t Find the value of k, if points (-2, 5), (-5, -10) and PRACTICE QUESTIONS between the coordinate axes by the line joining (k, -13) are collinear. the points (1, 2) and (3, 4) is ______. \t\t(a) 5 \t\t (b) \u221228 28 5 \t\t(a) 4\t\t (b) 6 \t\t(c) 28\t\t (d) 5 \t\t(c) 8\t\t (d) 2 4\t 7.\t The inclination of the line 3y \u2212 x + 24 = 0, is \t43.\t If A = (3, \u22124), B = (7, 0) and C = (14, \u22127) are ______. the three consecutive vertices of a parallelogram ABCD, then find the slope of the diagonal BD. \t\t(a) 60\u00b0\t\t (b) 30\u00b0 The following are the steps involved in solving the above problem. Arrange them in sequential order. \t\t(c) 45\u00b0\t\t (d) 135\u00b0 \t\t(A) \uf8eb x + 7 , y + 0\uf8f6 = \uf8eb 3 + 14 , \u22124 \u2212 7\uf8f6 . \t48.\t Find the product of intercepts of the line 3x + 8y \uf8ed\uf8ec 2 2 \uf8f7\uf8f8 \uf8ec\uf8ed 2 2 \uf8f8\uf8f7 - 24 = 0. \t\t(a) 8\t\t (b) 24 \t\t(B) The slope of BD = \u221211 \u2212 0 = \u221211 . \t\t(c) 3\t\t (d) 12 10 \u2212 7 3 \t49.\t Find the value of k, if points (10, 14), (\u22123, 3) and \t\t(C) x+7 = 17 and y + 0 = \u221211 (k, -8) are collinear. 2 2 2 2 \t\t(a) 16\t\t (b) 18 \u21d2 x = 10, y = \u221211 \t\t \t \u2234 D = (10, \u221211). \t\t(c) \u221218\t\t (d) \u221216","14.26 Chapter 14 \t50.\t The inclination of the line y \u2212 x + 11 = 0, is \t51.\t The equation of a line whose x-intercept is \u22123 and _______. which is parallel to 5x + 8y \u2212 7 = 0 is _______. \t\t(a) 30\u00b0 \t\t(a) 5x + 8y + 15 = 0 \t\t(b) 60\u00b0 \t\t(b) 5x + 8y \u2212 15 = 0 \t\t(c) 0\u00b0 \t\t(c) 5x + 8y \u2212 17 = 0 \t\t(d) 45\u00b0 \t\t(d) 5x \u2212 8y \u2212 18 = 0 Level 3 \t52.\t The area of a square with one of its vertices as \t\t(a) 12\t\t (b) 9 (5, -2) and the mid-point of the diagonals as (3, 2), \t\t(c) 6\t\t (d) 3 is _____. (in sq. units) 5\t 7.\t The equation of a line whose x-intercept is 11 and \t\t(a) 40\t\t (b) 20 perpendicular to 3x \u2212 8y + 4 = 0, is _______. \t\t(c) 60\t\t (d) 70 \t\t(a) 7x + 3y \u2212 77 = 0 \t\t(b) 8x + 3y \u2212 88 = 0 \t53.\t The equation of the line perpendicular to the line \t\t(c) 5x + 3y \u2212 55 = 0 inclined equally to the coordinate axes and passing \t\t(d) 3x + 8y \u2212 88 = 0 through (2, \u22123) is ______. 5\t 8.\t A(\u221211, 7) and B(\u221210, 6) are the points of trisection \t\t(a) x + y + 1 = 0 of a line segment PQ. Find the coordinates of P \t\t(b) x \u2212 y \u2212 2 = 0 and Q. \t\t(c) x + y + 2 = 0 \t\t(d) 2x + y \u2212 1 = 0 \t\t(a) (\u221212, 8); (\u22129, 5) \t\t(b) (\u221212, \u22128); (\u22129, 5) \t54.\t A triangle is formed by points (6, 0), (0, 0) and (0, \t\t(c) (12, 0); (9, \u22125) 6). How many points with the integer coordinates \t\t(d) (12, \u22128); (9, \u22125) are in the interior of the triangle? PRACTICE QUESTIONS \t\t(a) 7\t\t (b) 6 \t59.\t If one of the diagonals of a rhombus is 3x \u2212 4y + 10 = 0, then find the equation of the other diago- \t\t(c) 8\t\t (d) 10 nal which passes through point (\u22122, \u22123). 5\t 5.\t The equation of one of the diagonals of a square \t\t(a) 4x + 3y + 17 = 0 is 3x \u2212 8y + 4 = 0. Find the equation of the other \t\t(b) 3x \u2212 4y + 15 = 0 diagonal passing through the vertex (4, \u22126). \t\t(c) 4x + 3y \u2212 15 = 0 \t\t(d) 3x \u2212 4y \u2212 11 = 0 \t\t(a) 8x + 3y \u2212 15 = 0 \t\t(b) 3x \u2212 8y \u2212 11 = 0 \t60.\t The equation of the diagonal AC of a square \t\t(c) 8x + 3y \u2212 14 = 0 ABCD is 3x + 4y + 12 = 0. Find the equation of \t\t(d) 8x + 3y + 15 = 0 BD, where D is (2, \u22123). \t56.\t The lines 2x + 3y - 6 = 0 and 2x + 3y - 12 = 0 are \t\t(a) 4x \u2212 3y \u2212 8 = 0 represented on the graph. The difference between \t\t(b) 4x \u2212 3y \u2212 17 = 0 the areas of triangles formed by the lines with the \t\t(c) 4x \u2212 3y + 17 = 0 coordinate axes is _____. (in sq. units) \t\t(d) 4x + 3y \u2212 17 = 0","Coordinate Geometry 14.27 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t first 1\t 7.\t x = 19 2 \t2.\t (2, 3) \t3.\t \uf8ee\uf8f0\uf8ef3, \u2212 3\uf8f9 \t18.\t (-ve, +ve) 2 \uf8fb\uf8fa \t19.\t 0\u00ba \t4.\t right-angled isosceles triangle 2\t 0.\t \uf8eb 5 , 0\uf8f6\uf8f7\uf8f8 \uf8ed\uf8ec 3 \t5.\t straight line \t6.\t \u2212a 2\t 1.\t X-axis b \t22.\t a1a2 + b1b2 = 0 \t7.\t The point may lie in Q3 or Q4. 2\t 3.\t origin (or) (0, 0) \t8.\t x + 3 = 0 \t24.\t X \t9.\t 2 \t25.\t collinear 1\t 0.\t \u22122 \uf8eb \u2212c \uf8f6 \uf8ed\uf8ec b \uf8f8\uf8f7 1\t 1.\t 0 \t26.\t 0, 1\t 2.\t the first or third quadrant \t27.\t \u2212cot\u2009\u03b8 \t13.\t y = \u22122 x \u2212 5 \t28.\t fourth 3 3 2\t 9.\t 1 \t14.\t parallel \t30.\t parallel 1\t 5.\t collinear 1\t 6. 1 c2 sq. units \t 2 m Short Answer Type Questions \t31.\t 2x \u2212 3y + 13 = 0 3\t 9.\t x-intercept (a) = 2 3\t 2.\t 50\u03c0 sq. units \t\ty-intercept (b) = \u22123 \t33.\t \u03bb = \u22122 \t40.\t 30\u00b0 1 \u221213\uf8f6 \t34.\t k = \u221216 4\t 1.\t \uf8eb 14 , 14 \uf8f8\uf8f7 ANSWER KEYS \uf8ec\uf8ed 3\t 5.\t 85 units \t42.\t 3 3 x \u2212 3y \u2212 1= 0 4\t 3.\t (2, 0) 3\t 6.\t k= \u22125 \t45.\t 7x + 3y + 15 = 0 4 \t37.\t 20 sq. units \t38.\t 2 2a units Essay Type Questions \t46.\t x + y =1 or x + y =1 4\t 8.\t x \u2212 y + 16 = 0 1 1 \t49.\t 2x + 5y = 10 or 5x + 2y = 10 \u2212 1 \u2212 1 \t50.\t x + 3y \u2212 8 = 0 4 4 \t47.\t 4x \u22123y \u2212 2 = 0","14.28 Chapter 14 CONCEPT APPLICATION Level 1 \t1.\u2002(b)\t 2.\u2002 (d)\t 3.\u2002 (b)\t 4.\u2002 (d)\t 5.\u2002 (c)\t 6.\u2002 (a)\t 7.\u2002 (c)\t 8.\u2002 (d)\t 9.\u2002 (c)\t 10.\u2002 (b) \t11.\u2002 (a)\t 12.\u2002 (c)\t 13.\u2002 (a)\t 14.\u2002 (d)\t 15.\u2002 (c)\t 16.\u2002 (d)\t 17.\u2002 (b)\t 18.\u2002 (b)\t 19.\u2002 (b)\t 20.\u2002 (c) \t21.\u2002 (b)\t 22.\u2002 (d)\t 23.\u2002 (a)\t 24.\u2002 (b)\t 25.\u2002 (b)\t 26.\u2002 (a)\t 27.\u2002 (d)\t 28.\u2002 (a)\t 29.\u2002 (c)\t 30.\u2002 (d)\t Level 2 \t31.\u2002 (b)\t 32.\u2002 (c)\t 33.\u2002 (c)\t 34.\u2002 (d)\t 35.\u2002 (a)\t 36.\u2002 (c)\t 37.\u2002 (a)\t 38.\u2002 (d)\t 39.\u2002 (d)\t 40.\u2002 (d) \t41.\u2002 (c)\t 42.\u2002 (d)\t 43.\u2002 (c)\t 44.\u2002 (d)\t 45.\u2002 (d)\t 46.\u2002 (b)\t 47.\u2002 (b)\t 48.\u2002 (b)\t 49.\u2002 (d)\t 50.\u2002 (d) \t51.\u2002 (a) Level 3 54.\u2002 (d)\t 55.\u2002 (c)\t 56.\u2002 (b)\t 57.\u2002 (b)\t 58.\u2002 (a)\t 59.\u2002 (a)\t 60. (b) \t52.\u2002 (a)\t 53.\u2002 (a)\t ANSWER KEYS","Coordinate Geometry 14.29 CONCEPT APPLICATION Level 1 \t1.\t Fourth vertex of a parallelogram is (x1 + x3 - x2, y1 1\t 7.\t Identify the sign of y for each sign of x. + y3 - y2). \t18.\t Find lengths of the diagonals, then area of \t2.\t If x-intercept is a, then (a, 0) is a point on the line. Now, use two point form. rhombus = 1 \u00d7 d1 \u00d7 d2. 2 \t3.\t Use slope m = tan\u03b8, where q is the inclination. \t19.\t Find the distance between the given two points \t4.\t Equation of a line perpendicular to Y-axis is of the and compare that distance with the radius given. form x = constant. 2\t 0.\t The mid-point of the diameter is the centre of a \t5.\t (i) \u0007All the three vertices of an equilateral triangle circle. are not rational 2\t 1.\t Find the roots of the given equation, then use \t\t(ii) Let the third vertex be (x, y). intercepts form of line. \t\t(iii) Sides of an equilateral triangle are equal. \t22.\t (i) E\u0007 quation of a line parallel to X-axis is of the form y = constant. \t6.\t Find slopes of AB and BC, if they are equal then the given points are collinear. \t\t(ii) Slope of a line parallel to X-axis is zero. \t7.\t Equation of the line making equal intercepts is of \t\t(iii) Slope of a line = \u2212 (x-coefficient ) . the form, x + y = a. ( y -coeffcient ) \t8.\t The centre of a circle is the mid-point of the diameter. \t9.\t If x-intercept is 10, then the line passing through \t23.\t The slope of the line joining two points is y2 \u2212 y1 . (10, 0). Now, use the two point form. x2 \u2212 x1 1\t 0.\t Take the point on Y-axis as (0, y). 2\t 4.\t The first line is parallel to the Y-axis. The second Hints and Explanation \t11.\t Find the roots, and then use slope-intercept form line is parallel to the X-axis. of line. \t25.\t Two lines, which are perpendicular to the same 1\t 2.\t Area of square = (side)2. line, must be parallel to each other. 1\t 3.\t Use x + y =1 and a = \u2212b. 2\t 6.\t Solve the first two equations and substitute (x, y) a b in the third equation and evaluate k. 1\t 4.\t (i) \u0007Find the diameter of the circle then find its area. \t\t(ii) T\u0007 he distance between (\u22122, 3) and (6, \u22123) is the 2\t 7.\t Find the point of intersection of the given lines, length of the diameter of the circle. then decide. \t\t(iii)\tArea of circle \u03c0d 2 , where d is the length of 2\t 8.\t The circum-centre of a right-angled triangle is the 4 mid-point of its hypotenuse. = the diameter. \t29.\t Find the coordinates of the point of intersection. 1\t 5.\t Use m = tan\u2009q. \t30.\t The distance from centre of the circle to any point on the circle is its radius. 1\t 6.\t If two lines are perpendicular to the same line, then they are parallel to each other. Level 2 \t31.\t (i) \u0007P is the mid-point of the line joining the inter- \t\t(iii) \u0007Using the above data find a and b, then the cepts. Find the intercepts using the mid-point formula. equation of line, i.e., x + y = 1. a b \t\t(ii) \u0007Let the line cut coordinate axes at A(a, 0) and B(0, b). \t32.\t (i) Identify the type of \u0394OAB.","14.30 Chapter 14 \t\t(ii) \u0007Ortho-centre of right triangle is the vertex \t40.\t If (x1, y1), (x2, y2) and (x3, y3) are the successive containing right angle. vertices of a parallelogram, then the fourth vertex 3\t 3.\t (i) \u0007Find the roots and take the equation as = (x1 - x2 + x3, y1 - y2 + y3). x y = 1. (i) Find the roots, then use a b x y = 1. 4\t 1.\t + a b + \t\t(ii) Roots of x2\u2212 5x + 6 = 0 are 2 and 3. \t\t(ii) Roots of 3x2 \u2212 2x \u2212 1 = 0 are\u2009 \u22121 \u2009and 1. 3 \t\t(iii) x\u0007 -intercept is either 2 or 3. y-intercept is either \t\t(iii) x-intercept is either\u2009 \u22121 or 1. 3 or 2. 3 3\t 4.\t (i) Find m, then use slope-point form. \t\t \ty-intercept is either 1 or \u22121 . 3 \t\t(ii) \u0007Find the equation of the line passing through the given points. \t\t(iv) The required equation of the line is x + y = 1. a b \t\t(iii) \u0007Any line parallel to ax + by + c1 = 0 is ax + by + c2 = 0. 4\t 2.\t (i) \u0007Find the intercepts, then find the distance between the intercepted points. \t\t(iv) \u0007The required line ax + by + c2 = 0 passes through (5, 0). \t\t(ii) \u0007Find the equation of the line joining points (1, 2) and (3, 4). 3\t 5.\t (i) \u0007Area = 1 c2 , when the equation of the line is 2 ab \t\t(iii) F\u0007 ind the intercepts (a and b) of the line by put- ting x = 0 and y = 0. ax + by + c = 0. \t\t(ii) I\u0007 f a and b are x- and y-intercepts, then the area \t\t(iv) Length of the required line is a2 + b2 . of the triangle formed by the line with coordi- Hints and Explanation nate axes is\u2009 ab . 4\t 3.\t DACB is the required sequential order. 2 3\t 6.\t (i) The three points are collinear. \t44.\t CDBA is the required sequential order. \t\t(ii) Given points A, B and C are collinear. \t45.\t 7x \u2212 2y \u2212 14 = 0 \u21d2 7x \u2212 2y = 14 \t\t(iii) Use, slope of AB = slope of AC and find m. \t\t\u21d2 x \u2212 y =1\u21d2 x + y = 1. 2 7 2 (\u22127) 3\t 7.\t (i) F\u0007ind the intercepts, then find the distance between them. \t\t \u2234 \u0007Intercepts are 2 and \u22127, and their product = 2 \u00d7 (\u22127) = \u221214. \t\t(ii) \u0007Find the equation of the line joining the given points. \t46.\t (\u22122, 5), (\u22125, \u221210) and (k, \u221213) are collinear \t\t(iii) \u0007Find the intercepts (a and b) made by the above \u21d2 \u221210 \u2212 5 = \u221213 +10 line with coordinate axes. \u22125 + 2 k+5 \t\t(iv) The distance between (a, 0) and (0, b). \u221215 \u22123 \u22123 k+5 3\t 8.\t (i) Use the formula a(x - x1) + b(y - y1) = 0. \u21d2 = \t\t(ii) Slopes of parallel lines are equal. \u21d2 5(k + 5) = \u22123 \t\t(iii) \u0007Use, point-slope form, i.e., (y - y1) = m(x - \t\t x1) and find the equation of the line. \u21d2 k + 5 = \u22123 5 \t39.\t (i) The given vertices form an equilateral triangle. \u22123 \u221228 \t\t(ii) G\u0007 iven points are the vertices of an equilateral \u21d2 k = 5 \u2212 5 = 5 triangle. \t\t(iii) \u0007In any equilateral triangle, geometric centres \u2234 k = \u221228 . (except ex-centre) coincide. 5","Coordinate Geometry 14.31 \t47.\t 3y \u2212 x + 24 = 0 \u21d2 m = 1. \t\t\u21d2 \u221211 = \u221211 3 \u221213 k+3 \u2234m = tan\u03b8 = 1 \t\t \u21d2 k + 3 = \u221213 3 \t\t \u21d2 k = \u221216. \t\t\u21d2 tan\u03b8 = 1 5\t 1.\t y \u2212 x + 11 = 0 \u21d2 y = x \u2212 11 3 \t\t \u21d2 m = tan\u2009q = 1 \uf8ec\uf8ed\uf8eb\u2235 tan 30\u00b0 = 1\uf8f6 \u21d2 \u03b8 = 30\u00b0. 3 \uf8f7\uf8f8 \t\t \u21d2 \u03b8 = 45\u00b0. \u22125 8 \t48.\t 3x + 8y = 24 \t52.\t Slope of the line parallel to 5x + 8y \u2212 7 = 0 is . \t\t\u21d2 x y =1 \t\tGiven that the x-intercept of the required line 8 3 + is \u22123. \t\t \u21d2 I\u0007ntercepts are 8 and 3 and their product is 8 \u00d7 3 \t\t \u2234 I\u0007t passes through (\u22123, 0). Hence, the required = 24. line is \t49.\t Let the points be A = (10, 14), B = (-3, 3) and C \t\t\u2212 5 = x y\u22120 = (k, -8). 8 \u2212 ( \u22123) \t\tGiven, they are collinear \u21d2 The slopes are same. \t\ty \u2212 0 = \u22125 (x 3) 8 \t\t \u21d2 Slope of AB = Slope of BC + 3 \u2212 14 \u22128 \u2212 3 \t\t \u21d2 8y = \u22125x \u221215 \u21d2 5x + 8y + 15 = 0. \u22123 \u2212 10 k+3 \t\t\u21d2 = Level 3 Hints and Explanation \t53.\t (i) R\u0007 equired line is perpendicular to x = y and \t\t \u2234 It passes through (11, 0). passes through (2, -3). \t\tHence, the required line is \u2212 8 = y \u22120 . \t\t(ii) Find m, then use the slope-point form. 3 x \u2212 11 \t54.\t Draw the triangle and list the possible points. \t\tThat is, y \u2212 0 = \u2212 8 (x \u2212 11) 5\t 5.\t (i) I\u0007n a square, the diagonals are perpendicular to 3 each other. \t\t\u21d2 8x + 3y \u2212 88 = 0. \t\t(ii) \u0007Find the slope of the second diagonal and use 5\t 8.\t Let P = (a1, b1), and Q = (a2, b2). Given, A = (\u221211, the slope-point form. 7) and B = (\u221210, 6). \t56.\t (i) \u0007Find the intercepts made by the lines on the PQ coordinate axes by writing the equations in the AB intercept form. \t\t(ii) \u0007If the intercepts made are a and b, then the area \t\t \u21d2 A is the mid-point of the line segment PB. of the triangle formed is ab . 2 \t\t \u2234 A = \uf8eb a1 \u2212 10 , b1 + 6 \uf8f6 \uf8ed\uf8ec 2 2 \uf8f8\uf8f7 \t57.\t Slope of the line perpendicular to 3x \u2212 8y + 4 = 0 \t\t(\u221211, \uf8eb a1 \u2212 10 b1 + 6 \uf8f6 \u22121 \u22121 8 \uf8ec\uf8ed 2 2 \uf8f8\uf8f7 is Slope of 3x \u2212 8y 4 0 = 3 = \u2212 3 . 7) = , + = 8 \t\t \u21d2 a1 = \u221212 and b1 = 8 \t\tGiven that, x-intercept of the required line is 11. \t\t \u21d2 P = (\u221212, 8). Similarly, Q = (\u22129, 5).","14.32 Chapter 14 \t59.\t One diagonal of the rhombus is 3x \u2212 4y + 10 = 0. 6\t 0.\t The equation of diagonal AC is 3x + 4y +12 = 0. \t\tSlope of 3x \u2212 4y + 10 = 0 is 3 . \t\tIts slope =\u2212 3 . 4 4 \t\tIn a rhombus, diagonals bisect each other \t\tSlope of the other diagonal BD 1 = 4 . perpendicularly. \u22123 3 \t\t \u2234 The slope of the other diagonal 4 \t\t= \u22121 = \u2212 4 . \t\tBD passes through (2, \u22123) 3 3 4 y \u2212 (\u22123) 4 4 y \u2212 (\u22123) \u2234 3 = x\u22122 3 x \u2212 (\u22122) \t\tThe required equation is \u2212 = 4 3 4 \t\t\u2234 y + 3 = (x \u2212 2) \t\t\u21d2 y + 3 = \u2212 3 (x + 2) 4x \u2212 3y \u2212 17 = 0. \t\t\u21d2 4x + 3y + 17 = 0. Hints and Explanation","1152CChhaapptteerr lKoicnuesmatics REmEmBER Before beginning this chapter, you should be able to: \u2022 Understand the basic concepts of geometrical shapes and patterns \u2022 Know different types of angles and their values \u2022 Remember theorems related to triangles and angles KEy IDEaS After completing this chapter, you should be able to: \u2022 Understand the concept of locus and point of locus \u2022 Calculate the equation of a locus \u2022 Know the concept of congruency and learn the terms related to geometric centres of a triangle Figure 1.1","15.2 Chapter 15 INTRODUCTION Mark a fixed point O on a sheet of paper. Now, start marking points P1, P2, P3, P4, \u2026 on the sheet of paper such that OP1 = OP2 = OP3 = \u2026 = 4 cm. What do we observe on joining these points by a smooth curve? We observe a pattern, which is circular in shape and every point is at a distance of 4 cm from point O. It can be said that whenever points satisfying a certain condition are plotted, a pattern is formed. This pattern formed by all possible points satisfying the given condition is called the locus of points satisfying the given conditions. In the above example, we have the locus of points which are equidistant (4 cm) from the given point O. The collection (set) of all points which satisfy certain given geometrical conditions is called the locus of a point satisfying the given conditions. Alternatively, a locus can be defined as the path or curve traced by a point in a plane when subjected to some geometrical conditions. Consider the following examples: r 1.\t \u0007The locus of the point in a plane which is at a constant distance O r from a fixed point O is a circle with centre O and radius r units (see Fig. 15.1). 2.\t T\u0007 he locus of the point in a plane which is at a constant distance from a fixed straight line is a pair of lines, parallel to the fixed line. Let the fixed line be l. The lines, m and n, form the set of all points which are at a constant distance from l (see Fig. 15.2). m d Figure 15.1 l d n Figure 15.2 3. \t \u0007The locus of a point in a plane, which is d m equidistant from a given pair of parallel lines d l is a straight line, parallel to the two given lines n and lying midway between them. Figure 15.3 \t \u0007\t In Fig. 15.3, m and n are the given lines and line l is the locus. Before proving that a given path or curve is the desired locus, it is necessary to prove the following: 1.\t Every point lying on the path satisfies the given geometrical conditions. 2.\t Every point that satisfies the given conditions lies on the path. Example 15.1 Show that the locus of a point, equidistant from the endpoints of a line segment, is the perpendicular bisector of the segment. Solution The proof will be taken up in two steps.","Locus 15.3 Step 1:\u2002 We initially prove that any point equidistant from A the endpoints of a line segment lies on the perpendicular bisector of the line segment. Given: M and N are two points on a plane. A is a point in the same plane such that AM = AN (see Fig. 15.4). To prove: A lies on the perpendicular bisector of MN. Proof: Let L be the mid-point of MN. If A coincides with L, then A lies on the bisector of MN. M L N N Suppose A is different from L. Then, in \u0394MLA and \u0394NLA, ML = NL, AM = AN and AL is a common side. Figure 15.4 \u2234 By SSS congruence property, \u0394MLA \u2245 \u0394NLA. \u21d2 \u2220MLA = \u2220NLA ( The corresponding elements of congruent triangles are equal.)\b (1) But, \u2220MLA + \u2220NLA = 180\u00b0 ( \u2009Linear pair) l \u21d2 2 \u2220MLA = 180\u00b0 (From Eq. (1)) \u2234 \u2220MLA = \u2220NLA = 90\u00b0. P So, AL \u22a5 MN. Hence, AL is the perpendicular bisector of MN. \u2234 A lies on the perpendicular bisector of MN. Step 2:\u2002 Now, we prove that any point on the perpendicular || || bisector of the line segment is equidistant from the end points L of the line segment. M Given: MN is a line segment and P is a point on the perpendicular bisector. L is the mid-point of MN (see Fig. 15.5). To prove: MP = NP. Proof: If P coincides with L, then MP = NP. Figure 15.5 Suppose P is different from L. Then, in \u0394MLP and \u0394NLP, ML = LN. LP is the common side and \u0394MLP = \u0394NLP = 90\u00b0. \u2234 By the SAS congruence property, \u0394MLP \u2245 \u0394NLP. So, MP = PN ( The corresponding elements of congruent triangles are equal.) \u2234 Any point on the perpendicular bisector of MN is equidistant from points M and N. Hence, from Step I and Step II of the proof, it can be said that the locus of point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points. Example 15.2 Show that the locus of a point, equidistant from two intersecting lines in the plane, is a pair of lines bisecting the angles formed by the given lines. Solution Step 1: We initially prove that any point, equidistant from two given intersecting lines, lies on one of the lines bisecting the angles formed by the given lines.","15.4 Chapter 15 Given: AB and CD are two lines intersecting at C m B O. P is the point on the plane such that PM = PN. M Line l is the bisector of \u2220BOD and \u2220AOC. OP Line m is the bisector of \u2220BOC and \u2220AOD (see Fig. 15.6). To prove: P lies either on the line l or on the line m. Proof: In \u0394POM and \u0394PON, PM = PN, N OP is a common side and \u2220PMO = \u2220PNO = 90\u00b0. A D \u2234 By RHS congruence property, \u0394POM \u2245 \u0394PON. So, \u2220POM = \u2220PON, i.e., P lies on the angle Figure 15.6 bisector of \u2220BOD. As l is the bisector of \u2220BOD and \u2220AOC, P lies on the line l. Similarly, if P lies in any of the regions of \u2220BOC, \u2220AOC or \u2220AOD, such that it is equidistant from AB and CD, then we can conclude that P lies on the angle bisector l or on the angle bisector m. Step 2:\u2002 Now, we prove that any point on the bisector of one m of the angles formed by two intersecting lines is equidistant from the lines. C MB Given: Lines AB and CD, intersect at O. Lines l and m are the O P l angle bisectors. N AD Proof: Let l be the angle bisector of \u2220BOD and \u2220AOC, and m be the angle bisector of \u2220BOC and \u2220AOD. Let P be a point on the angle bisector l, as shown in Fig.\u00a015.7. Figure 15.7 If P coincides with O, then P is equidistant from line AB and CD. Suppose P is different from O. Draw the perpendiculars PM and PN from the point P onto the lines AB and CD respectively. Then in \u0394POM and \u0394PON, \u2220POM = \u2220PON, \u2220PNO = \u2220PMO = 90\u00b0 and OP is a common side. \u2234 By the AAS congruence property, \u0394POM \u2245 \u0394PON. So, PN = PM ( The corresponding sides in congruent triangles.) That is, P is equidistant from lines AB and CD. Hence, from Step I and Step II of the proof, it can be said that the locus of the point, which is equidistant from the two intersecting lines is the pair of the angle bisectors of the two pairs of vertically opposite angles formed by the lines. Equation of a Locus We know that the locus is the set of points that satisfy a given geometrical condition. When we express the geometrical condition in the form of an algebraic equation, the equation is called the equation of the locus.","Locus 15.5 Steps to Find the Equation of a Locus 1. \t Consider any point (x1, y1) on the locus 2.\t Express the given geometrical condition in the form of an equation using x1 and y1. 3.\t Simplify the equation obtained in Step 2. 4.\t \u0007Replace (x1, y1) by (x, y) in the simplified equation, which gives the required equation of the locus. The following formulae will be helpful in finding the equation of a locus: 1.\t Distance between two points A(x1, y1) and B(x2, y2) is AB = (x2 \u2212 x1 )2 + (y2 \u2212 y1 )2 . 2.\t Area of the triangle formed by joining points (x1, y1), (x2, y2) and (x3, y3) is \t 1 x1 \u2212 x2 x2 \u2212 x3 , where the value of a b = ad \u2212 bc. 2 y1 \u2212 y2 y2 \u2212 y3 c d 3.\t Equation of the circle with centre (a, b) and radius r is given by (x \u2212 a)2 + (y \u2212 b)2 = r2. 4.\t The perpendicular distance from a point P(x1, y1) to a given line ax + by + c = 0 is ax1 + by1 + c . a2 + b2 Example 15.3 Find the equation of the locus of a point which forms a triangle of area 5 square units with the points A(2, 3) and B(\u22121, 4). Solution Let P(x1, y1) be point on the locus, (x2, y2) = (2, 3) and (x3, y3) = (\u22121, 4). Given area of \u0394PAB = 5 sq. units. \u2234 1 x1 \u22122 2 \u2212 (\u22121) =5 2 y1 \u22123 3\u22124 \u21d2 x1 \u2212 2 3 = 10 \u21d2 \u2212 (x1 \u2212 2) \u2212 3(y1 \u2212 3) = \u00b110 y1 \u2212 3 \u22121 \u21d2 x1 + 3y1 \u2212 11 = \u00b110. The required equation is, \tx + 3y = 10 + 11 \t or\t x + 3y = \u221210 + 11 \tx + 3y \u2212 21 = 0\t or \t x + 3y \u2212 1 = 0 Concurrency\u2014Geometric Centres of a Triangle Let us recall that if three or more lines pass through a fixed point, then the lines are said to be concurrent and the fixed point is called the point of concurrence. In this context, we recall different concurrent lines and their points of concurrence associated with a triangle. Geometric Centres of a Triangle Circum-centre The locus of the point equidistant from the endpoints of the line segment is the perpendicular bisector of the line segment. The three perpendicular bisectors of the three sides of a triangle","15.6 Chapter 15 are concurrent. The point of their concurrence is called the A circum-centre of the triangle. It is usually denoted by S. The circum-centre is equidistant from all the vertices of the triangle. R SQ The circum-centre of the triangle is the locus of the point in the B PC plane of the triangle equidistant from the vertices of the triangle (see Fig. 15.8). Figure 15.8 A In-centre I The angle bisectors of the triangle are concurrent. The point of concurrence is called in-centre. It is usually denoted by I. The BC point I is equidistant from the sides of the triangle. The in- Figure 15.9 centre of the triangle is the locus of the point, in the plane of the A triangle, equidistant from the sides of the triangle (see Fig. 15.9). BC Ex-centre The point of concurrence of the external bisector of two angles of a triangle and the internal bisector of the third angle is called ex-centre of the triangle (see Fig.\u00a015.10). Ortho-centre The altitudes of the triangle are concurrent and the point of concurrence of the altitudes of a triangle is called ortho-centre. It is usually denoted by O (see Fig. 15.11). A E I1 Figure 15.10 C O BD C Figure 15.11 Centroid The medians of a triangle are concurrent and the point of concurrence of the medians of a triangle is called centroid. It is usually denoted by G. The centroid divides each of the medians in the ratio 2 : 1, beginning from vertex, i.e., AG : GD = 2 : 1, BG : GE = 2 : 1 and CG : GF = 2 : 1, as shown in Fig. 15.12. A E FG BD C Figure 15.12","Locus 15.7 Some Important Points 1.\t \u0007In an equilateral triangle, the centroid, the ortho-centre, the circum-centre and the in- centre all coincide. 2.\t I\u0007n an isosceles triangle, the centroid, the ortho-centre, the circum-centre and the in- centre all lie on the median to the base. 3.\t I\u0007n a right triangle, the length of the median drawn to the hypotenuse is equal to half of the hypotenuse. The median is also equal to the circum-radius. The mid-point of the hypotenuse is the circum-centre. 4.\t \u0007In an obtuse-angled triangle, the circum-centre and ortho-centre lie outside the triangle. For an acute-angled triangle, the circum-centre and the ortho-centre lie inside the triangle. 5.\t F\u0007 or all triangles, the centroid and the in-centre lie inside the triangle. 6.\t F\u0007 or all triangles, the ex-centre lies outside the triangle.","15.8 Chapter 15 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t Centroid divides the median from the vertex in \t8.\t The path traced out by a moving point which the ratio _____. moves according to some given geometrical con- ditions is _____. \t2.\t Any point on the perpendicular bisector of a line segment joining two points is _____ from the two \t9.\t Is the statement \u2018In \u0394ABC, a point equidistant points. from AB and AC lies on the median\u2019, true? \t3.\t What is the locus of a point in a plane which is at a 1\t 0.\t The path of a freely falling stone is _____. distance of p units from the circle of radius q units (p \u2260 q)? \t11.\t The orthocentre of a right triangle is the _____. \t4.\t If A and B are two fixed points, then the locus of a 1\t 2.\t The line segment joining the vertex of a triangle point P, such that \u2220APB = 90\u00b0 is _____. and the mid-point of its opposite side is called _____. \t5.\t The locus of the point equidistant (in a plane) from the three vertices of a triangle is _____. 1\t 3.\t The locus of the centre of the circles (in a plane) passing through two given points is the _____ of \t6.\t The locus of the point in a plane which is the line segment joining the two points. equidistant from two intersecting lines is _____. \t14.\t The circum-centre of a right triangle always lies \t7.\t The line segment from the vertex of a triangle _____. perpendicular to its opposite side is called _____. 1\t 5.\t The locus of the tip of the hour hand is _____. Short Answer Type Questions \t16.\t Find the locus of the vertex of a triangle with fixed A base and having constant area. D PRACTICE QUESTIONS 1\t 7.\t Find the locus of the centre of a circle passing through two fixed points A and\u00a0B. \t18.\t Let A and B be two fixed points in a plane. Find BC the locus of a point P, such that PA2 + PB2 = AB2. \t22.\t Show that the locus of a point, equidistant from \t19.\t Find the locus of the point P, such that TP\u00a0:\u00a0MP = three distinct given collinear points in a plane does 3 : 2, where T is (\u22122, 3) and M is (4, \u22125). not exist. \t20.\t Find the locus of the point which is equidistant 2\t 3.\t If ZQ and RU be two lines intersecting at point E, from sides AB and AD of a rhombus ABCD. then find the locus of a point moving in the inte- rior of \u2220UEZ, such that the sum of its distances 2\t 1.\t In the following figure (not to scale), ABC is a from the lines ZQ and RU is b units. right isosceles triangle, right-angled at B and BD \u22a5 AC. If the triangle ABC is rotated about 2\t 4.\t Find the locus of the point which is equidistant the hypotenuse, then find the locus of the triangle from the sides AB and AC of triangle ABC. ABC. 2\t 5.\t \u0394APQ, \u0394BPQ and \u0394CPQ are three isosceles trian- gles with the same base PQ. Show that the points A, B and C are collinear. Essay Type Questions \t26.\t In a square ABCD, if A and B are (5, 1) and (7, 2\t 7.\tIf P(x, y) and Q(1, 4) are the points on the 1) respectively, then what is the locus of the mid- circle whose centre is C(5, 7), then find the point of diagonal AC? locus of P.","Locus 15.9 \t28.\t If two lines intersect at P at right angles and pass 8 sq. units. If A is (2, 5) and B is (3, 4), then what through A(1, 1) and B(1, 0) respectively then what is the locus of P? is the locus of P? \t30.\t If P is a point on the circle with AB as a diameter, \t29.\t In \u0394PAB, D, E and F are the mid-points of PA, where A and B are (0, 2) and (2, 4) respectively, AB and BP, respectively. The area of DEF is then the locus of P is _____. CONCEPT APPLICATION Level 1 \t1.\t The locus of a point equidistant from three fixed \t\t(a) x = \u00b1k\t\t (b) y = \u00b1k points is a single point. The three points are _____. \t\t(c) x = 0\t\t (d) y = 0 \t\t(a) collinear\t\t (b) non-collinear \t8.\t The locus of the centre of a wheel rolling on a straight road is a _____. \t\t(c) coincidental\t (d) None of these \t2.\t The locus of a point moving in a space which is at \t\t(a) concentric circle\t (b) straight line a constant distance from a fixed point in space is \t\t(c) curve path\t (d) parabola called a _____. \t9.\t If A and B are two fixed points, then the locus \t\t(a)\t square\t\t (b) sphere of a point P, such that PA2 + PB2 = AB2 is a\/an \t\t(c) circle\t\t (d) triangle ______. \t3.\t The locus of the centre of the circle that touch a \t\t(a) circle with AB as the diameter given circle internally is a _____. \t\t(b) right triangle with \u2220P = 90\u00b0 \t\t(a) straight line\t (b) hellix \t\t(c) semi-circle with AB as the diameter \t\t(c) circle\t\t (d) None of these \t\t(d) c\u0007 ircle with AB as the diameter, excluding \t4. \tIn \u0394ABC, \u2220A = \u2220B + \u2220C, then the circum- points A and B centre is at _____. 1\t 0.\t In the figure, RX \u22a5 PA, RY \u22a5 PB, RX = RY and PRACTICE QUESTIONS \t\t(a) A\t\t (b) B \u2220APB = 70\u00b0. Find \u2220APQ. \t\t(c) C\t\t (d) the mid-point of BC \t\t(a) 70\u00b0\t\t (b) 140\u00b0 \t5.\t The locus of a point equidistant from three points \t\t(c) 35\u00b0\t\t (d) 50\u00b0 does not exist. This implies that the three points are _____. X A R \t\t(a) collinear\t\t (b) non-collinear P Y \t\t(c) coincidental\t (d) None of these Q \t6.\t The locus of a point which is equidistant from two B non-intersecting lines l and m is a _____. \t11.\t Consider a point M inside a quadrilateral PQRS. \t\t(a) straight line parallel to the line l If M be the point of intersection of angle bisectors \t\t(b) straight line parallel to the line m PE and QF, then _____. \t\t(c) \u0007straight line parallel to the lines l and m and \t\t(a) M is nearer to PS than to QR midway between them \t\t(b) M is equidistant from opposite sides PS andQR \t\t(d) \u0007straight line that intersects both the lines l and\u00a0m \t\t(c) M is nearer to QR than to PS \t\t(d) M is equidistant from opposite sides\u2009PQ and SR \t7.\t The locus of a point which is at a constant distance k from Y-axis is _____.","15.10 Chapter 15 PQ 1\t 4.\t The locus of a point which is equidistant from the coordinate axes can be a _____. M FE \t\t(a) line making a non-zero intercept on X-axis SR \t\t(b) line making a non-zero intercept on Y-axis \t\t(c) \u0007line passing through the origin making an angle \t12.\t The locus of the point which is equidistant from the three lines determined by the sides of a triangle of 45\u00b0 with X-axis is _____. \t\t(d) None of the above \t\t(a) the in-centre \t15.\t The locus of a point equidistant from two inter- secting lines PQ and RS, and at a distance of 10 cm \t\t(b) the ex-centre from their point of intersection O is _____. \t\t(c) the ortho-centre \t\t(d) either (a) or (b) \t\t(a) f\u0007our points lying on the angle bisectors at a distance of 5 cm from O \t13.\t Find the locus of any fixed point on the circumfer- ence of a coin when the coin is rolling on a straight \t\t(b) \u0007two points lying on the angle bisectors at a dis- path. tance of 10 cm from O \t\t(a) Circle\t\t (b) Straight line \t\t(c) f\u0007our points lying on the angle bisectors at a \t\t(c) Sphere\t\t (d) Helix distance of 10 cm from O \t\t(d) \u0007two points lying on the angle bisectors at a dis- tance of 5 cm from O Level 2 1\t 6.\t In the figure following figure, O is the centre of the A circle and AL \u22a5 MN. If \u2220AOB = 90\u00b0, then find \u2220AOP. PRACTICE QUESTIONS \t\t(a) 60\u00b0\t\t (b) 20\u00b0 BC \t\t(c) 30\u00b0\t\t (d) 45\u00b0 D O \t19.\t A part of the locus of a point P, which is equidis- A PB tant from two intersecting lines ax + by + c = 0 and px + qy + r = 0, is _____. \t17.\t The solid formed when a right triangle is rotated about one of the sides containing the right angle is \t\t(a) (a \u2212 p)x + (b \u2212 q)y + (c \u2212 x) = 0 a _____. \t\t(b) apx + qby + cy = 0 \t\t(a) prism\t\t (b) cylinder \t\t(c) a2 + b2 ( px + qy + r ) = p2 + q2 (ax + by + c ) \t\t(c) cone\t\t (d) sphere \t \t (d) None of these \t18.\t In the figure (not to scale), AB = AC and BD = CD. Find \u2220ADB. \t20.\t If PAB is a triangle of area 4 sq. units and A is (2, 5) and B is (3, 4), then part of the locus of P is _____. \t\t(a) 60\u00b0\t\t (b) 90\u00b0 \t\t(c) 120\u00b0\t\t (d) Cannot be determined \t\t(a) x \u2212 y + 15 = 0 \t\t(b) x \u2212 y \u2212 15 = 0 \t\t(c) x + y \u2212 15 = 0 \t\t(d) x + y + 15 = 0 \t21.\t Given, triangle PBC and parallelogram ABCD lie between the same parallel lines. On the same base, BC and the area of parallelogram is 2 sq. units.","Locus 15.11 The points B and C are (2, 4) and (4, 4) respectively. 2\t 6.\t The locus of the centre of a circle that touches the Which of the following lines is a part of the locus given circle externally is a _____. of P? \t\t(a) curve\t\t (b) straight line \t\t(a) y = 4\t\t (b) y = 7 \t\t(c) circle\t\t (d) helix. \t\t(c) y = 5\t\t (d) y = 6 \t27.\t The locus of a rectangle, when the rectangle is \t22.\t If PAB is a triangle in which \u2220B = 90\u00b0 and A(1, 1) rotated about one of its sides, is a _____. and B(0, 1), then the locus of P is _____. \t\t(a) plane\t\t (b) sphere \t\t(a) y = 0\t\t (b) xy = 0 \t\t(c) cone\t\t (d) cylinder. \t\t(c) x = y\t\t (d) x = 0 2\t 8.\t If the ortho-centre of a triangle ABC is B, then 2\t 3.\t The locus of P whose distance from the X-axis is which of the following is true? thrice the distance from the line x = 5 is _____. \t\t(a) AC2 = AB2 + BC2 \t\t(a) x \u2212 y \u2212 5 = 0 \t\t(b) AC2 > AB2 + BC2 \t\t(b) 3x \u2212 y \u2212 15 = 0 \t\t(c) AC2 < AB2 + BC2 \t\t(c) 3x + y + 15 = 0 \t\t(d) None of these \t\t(d) x + y + 5 = 0 \t24.\t Two of the vertices of a triangle ABC are A(1, 1), \t29.\t In \u0394ABC, \u2220A = \u2220B + \u2220C. The point which is B(\u22121, \u22123) and the area of \u0394ABC is 6 sq. units. If P is equidistant from A, B and C is _____. the centroid of the \u0394ABC, then find the locus of P. \t\t(a) mid-point of AB\t \t\t(a) 2x \u2212 y + 1 = 0\t (b) 2x \u2212 y \u2212 3 = 0 \t\t(b) mid-point of AC \t\t(c) 2x + y + 3 = 0\t (d) Both (a) and (b) \t\t(c) mid-point of BC \t\t(d) None of these \t25.\t In a circle with radius 25 cm, what is the area of the region determined by the locus of the mid- 3\t 0.\t The locus of a point which is equidistant from points of chords of length 48 cm? (0, 2) and (0, 8) is _____. \t\t(a) 154 cm2\t\t (b) 254 cm2 \t\t(a) y = 4\t\t (b) y = 5 PRACTICE QUESTIONS \t\t(c) 72 cm2\t\t (d) None of these \t\t(c) x = 4\t\t (d) x = 5 Level 3 \t31.\t The area of \u0394PQR is 4 sq. units Q and R are (1, \t\t(a) median\t\t (b) centroid 1) and (1, 0), respectively. Which of the following \t\t(c) incentre\t\t (d) angle bisector lines is a part of the locus of P? \t34.\t The locus of a point which is collinear with the \t\t(a) x \u2212 6 = 0\t (b) x \u2212 7 = 0 two given points is _____. \t\t(c) x + 8 = 0\t (d) x + 7 = 0 \t\t(a) a circle\t\t (b) a triangle \t32.\t What is the locus of the point P(x, y) (where \t\t(c) a straight line\t (d) a parabola xy > 0), which is at a distance of 2 units from the origin? 3\t 5.\t The locus of a point, which is at a distance of 8 units from (0, \u22127), is _____. \t\t(a) x2 + y2 = 4 \t\t(b) x2 + y2 = 4, x > 0, y > 0 \t\t(a) x2 + y2 + 6x + 14y \u2212 15 = 0 \t\t(c) x2 + y2 = 4, x < 0, y < 0 \t\t(b) x2 + y2 + 14y \u2212 15 = 0 \t\t(c) y2 + 14y \u2212 8 = 0 \t\t(d) None of these \t\t(d) x2 + y2 + 14x + 14y \u2212 15 = 0 \t33.\t The locus of a point which is twice as far from \t36.\t The area of \u0394ABC is 2 sq. units. If A = (2, 4) and each vertex of a triangle as it is from the mid-point B = (4, 4), then find the locus of C(x, y). of the opposite side is a\/an\/the ______.","15.12 Chapter 15 \t\t(a) y \u2212 6 = 0 \t41.\tO is an interior point of a rhombus, ABCD, and \t\t(b) y \u2212 2 = 0 O is equidistant from BC and CD. Then O lies \t\t(c) Both (a) and (b) on _____. \t\t(d) None of these \t\t(a) AC 3\t 7.\t Find the locus of a point which is a constant dis- tance of 4 units away from the point (2, 4). \t\t(b) BD \t\t(c) Either (a) or (b) \t\t(a) x2 + y2 \u2212 4x \u2212 8y + 4 = 0 \t\t(d) Neither (a) nor (b) \t\t(b) x2 + 4x + 16 = 0 \t\t(c) y2 \u2212 8y + 12 = 0 \t42.\t In a triangle ABC, D is a point on BC, such that \t\t(d) None of these any point on AD is equidistant from points B and C. Which of the following is necessarily true? 3\t 8.\t P is the point of intersection of the diagonals of a square READ. P is equidistant from ______. \t\t(a) AB = BC \t\t(a) the vertices R, E, A and D \t \t (b) BC = AC \t\t(b) RE and EA \t\t(c) EA and AD \t\t(c) AC = AB \t\t(d) All of these \t \t (d) AB = BC = AC 3\t 9.\t A coin of radius 1 cm is moving along the circum- 4\t 3.\t The locus of a point, equidistant from the coordi- ference and interior of a square of side 5 cm. Find nate axes is _____. the locus of centre of the coin. \t\t(a) x = |y| \t\t(a) A square of side 6 cm \t\t(b) A square of side 4 cm \t\t(b) y = |x| \t\t(c) A square of side 3 cm \t\t(d) A square of side 2 cm \t\t(c) Both (a) and (b) \t40.\t ABC is a triangle in which AB = 40 cm, BC = 41 \t\t(d) None of these cm and AC = 9 cm. Then ortho-centre of \u0394ABC PRACTICE QUESTIONS lies ______. 4\t 4.\t P is an interior point of an equilateral triangle ABC. If P is equidistant from AB and BC, BC and \t\t(a) interior of the triangle AC, then \u2220BPC = _____. \t\t(b) exterior of the triangle \t\t(a) 120\u00b0\t\t (b) 90\u00b0 \t\t(c) on the triangle \t\t(c) 60\u00b0\t\t (d) 150\u00b0 \t\t(d) at the mid-point of the triangle 4\t 5.\t The locus of a point, which is equidistant from (2, 6) and (2, 8), is _____. \t\t(a) y = 7\t\t (b) x = 7 \t\t(c) x = 2\t\t (d) y = 2","Locus 15.13 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 2 : 1 \t7.\t altitude \t8.\t locus \t2.\t equidistant \t9.\t No \t10.\t vertical line \t3.\t Another circle of radius (p + q) or (q \u2212 p) concen- \t11.\t vertex containing the right angle tric to it. 1\t 2.\t median \t13.\t perpendicular bisector \t4.\t circle with diameter AB excluding the points A 1\t 4.\t on the mid-point of the hypotenuse and B. \t15.\t circle \t5.\t circum-centre \t6.\t angle bisectors of the two pairs of vertically oppo- site angles of intersecting lines. Short Answer Type Questions \t16.\t The required locus is a line parallel to the given 2\t 0.\t The required locus is diagonal AC. base. 2\t 1.\tAccording to given conditions, two equal cones \t17.\t The required locus is the perpendicular bisector of of base radius BD and slant height AB or BC are AB, except the mid-point of AB. formed in such a way that their bases touch one another. \t18.\t The required locus is the circle with diameter AB, excluding points A and B. \t23.\t The required locus is the angular bisector of the angle RPM. \t19.\t 5x2 + 5y2 \u2212 88x + 114y + 317 = 0 is the required locus of point P. \t24. \tThe required locus is the bisector of \u2220BAC. Essay Type Questions \t26.\t y = (2, 0) 2\t 9.\tx + y \u2212 71 = 0; x + y + 57 = 0 \t27.\t x2 + y2 \u2212 10x \u2212 14y + 49 = 0 \t30.\t x2 + y2 \u2212 2x \u2212 6y + 8 = 0 2\t 8.\t x2 + y2 \u2212 2x \u2212 y + 1 = 0 CONCEPT APPLICATION 5.\u2002 (a) \t 6.\u2002(c)\t 7.\u2002(a)\t 8.\u2002(b)\t 9.\u2002(d)\t 10.\u2002(c) ANSWER KEYS 15.\u2002(c) Level 1 \t 1.\u2002 (b) \t 2.\u2002(b)\t 3.\u2002 (c) \t 4.\u2002(d)\t \t11.\u2002(b)\t 12.\u2002(d)\t 13.\u2002(d)\t 14.\u2002(c)\t Level 2 18.\u2002(b)\t 19.\u2002(c)\t 20.\u2002(c)\t 21.\u2002(c)\t 22.\u2002(d)\t 23.\u2002(b)\t 24.\u2002(d)\t 25.\u2002(a) 28.\u2002(a)\t 29.\u2002(c)\t 30.\u2002(b) \t16.\u2002(d)\t 17.\u2002(c)\t \t26.\u2002(c)\t 27.\u2002(d)\t Level 3 33.\u2002(b)\t 34.\u2002(c)\t 35.\u2002(b)\t 36.\u2002(c)\t 37.\u2002(a)\t 38.\u2002(d)\t 39.\u2002(c)\t 40.\u2002(c) 43.\u2002(c)\t 44.\u2002(a)\t 45.\u2002(a) \t31.\u2002(d)\t 32.\u2002(d)\t \t41.\u2002(a)\t 42.\u2002(c)","15.14 Chapter 15 Concept Application Level 1 \t1.\t Three points form a triangle. \t8.\t When a wheel is rolling on the straight path, the distance from centre to the path remains same. \t2.\t Recall the definition. \t9.\t Angle in semicircle is 90\u00b0. \t4.\t Recall the properties of right triangle with respect to geometric centres. 1\t 0.\t AB is bisector of \u2220APB. 1\t 2.\t Recall the definitions of geometric centres. \t6.\t Non-intersecting lines are parallel. \t14.\t Required locus is a line which is in the midway of \t7.\t Equations of the lines which are k units from the X-axis and Y-axis. Y-axis. 1\t 5.\t Points on angle bisectors. Level 2 \t17.\t (i)\tWhen a right triangle is rotated about one 2\t 2.\t (i) Angle in a semicircle is 90\u00b0. of its perpendicular sides, the other perpen- \t\t(ii) \u0007Use, (PA)2 = (PB)2 + (AB)2 and obtain the dicular side acts as radius of the base and the hypotenuse acts as the slant height of solid. required locus. \t\t(ii)\tThe top of a solid is a point (vertex). 2\t 3.\t (i) \u0007The perpendicular distance from P(x1, y1) to \t18.\t \u0394ADB \u2245 \u0394ADC. the line ax + by + c = 0 is 1\t 9.\t (i) P\u0007 erpendicular distance of a point P(x1, y1) from ax1 + by1 + c . the line, px + qy + r = 0, is a2 + b2 Hints and Explanation px1 + qy1 + r . \t\t(ii) T\u0007 he distance between P and X-axis is equal to p2 + q2 thrice the distance between P and x = 5. \t\t(ii) P\u0007 erpendicular distance of the point (x1, y1) to \t24.\t (i) \u0007As P is the centroid of \u0394ABC, area of \u0394PAB the line ax + by + c = 0 is 1 is 3 (area of \u0394ABC\u2009). ax1 + by1 + c . a2 + b2 \t\t(ii) Area of \u0394PAB is given by 2\t 0.\t (i) Given, area of a triangle ABC = 4 1 x1 \u2212 x2 x2 \u2212 x3 . 2 y1 \u2212 y2 y2 \u2212 y3 \u21d2 1 x1 \u2212 x2 x2 \u2212 x3 = 4. \t25.\t (i) \u0007Locus of the mid-points of equal chords in a 2 y1 \u2212 y2 y2 \u2212 y3 circle forms a concentric circle with the given circle. \t\t(ii) S\u0007 ubstitute (x2, y2) = (2, 5) and (x3, y3) = (3, 4) and obtain the relation. \t\t(ii) The required locus is a circle of radius 7 cm. \t\t(iii) Now, find its area. \t21.\t \u2009\u2009(i) \u0007The area of the given triangle will be half that \t26.\tThe required locus is a circle. of the given parallelogram. 2\t 7.\t The required locus is a cylinder. \t\t\u2009\u2009(ii) \u0007Area of \u0394PBC is half of the area of parallelo- gram ABCD. 2\t 8.\t In a right triangle, the ortho- centre lies at the vertex which \t\t(iii) Area of \u0394PBC is given by contains right angle. 1 x1 \u2212 x2 x2 \u2212 x3 \t\t\u2234 \u0007The given triangle is a right triangle, where \u2220B 2 y1 \u2212 y2 y2 \u2212 y3 = 90\u00b0. \t\t\t and proceed same as above. \t \u2234 AC2 = AB2 + BC2.","Locus 15.15 \t29.\t Given, in a \u0394ABC, \u2220A = \u2220B + \u2220C \t\t\u2234 The required locus is y = 5. y=5 X-axis \u21d2 \u2220A= 90\u00b0, and BC is the hypotenuse. Y-axis (0, 8) \t\tThe point equidistant from the vertices of a triangle is circum-centre of the triangle. (0, 2) O \t\tFor a right triangle, circum-centre lies at the mid- point of the hypotenuse. \t\t\u2234 The required point is the mid-point of BC. 3\t 0.\t The locus of a point which is equidistant from two points is the perpendicular bisector of the line seg- ment joining the points. Level 3 \t31.\t (i) Area of triangle is given by \t\t\u2234 P is equidistant from the vertices. \t\tAs P lies on the diagonals, P is equidistant from 1 x1 \u2212 x2 x2 \u2212 x3 . 2 y1 \u2212 y2 y2 \u2212 y3 any two adjacent sides. \t\t\u2234 Hence, the correct answer is option (d). \t\t(ii) \u0007Substitute (x2, y2) = (1, 1), (x3, y3) = (1, 0) and \t39.\t When a coin is moving on the circumference of obtain the relation in terms of x1 and y1. a square, the path of the center of the coin is a \t32.\t (i) If xy > 0, then x > 0, y > 0 (or) x < 0, y < 0. square. \t\t(ii) Given, OP = 2, where O = (0, 0) and P = (x, y). \t\t\u2234 T\u0007 he required locus is a square of side (5 \u2212 2) cm, \t\t(iii) Substitute the values and obtain the locus. i.e., 3 cm. 3\t 3.\t (i) Recall the concept of medians in a triangle. Hints and Explanation \t\t(ii) \u0007The point of concurrence of medians is the \t40.\t Given, AB = 40 cm, BC = 41 cm and AC = 9 cm. \t\t\u2234 ABC is right triangle, right angled at A. centroid. \t\t\u2234 Ortho-centre lies at A. \t\t(iii) Centroid divides each median in the ratio 2 : 1. \t\t\u2234 Hence, the correct answer is option (c). 4\t 1.\t Given, O is equidistant from BC and CD. \t36.\t Area of A(2, 4), B(4, 4) and C(x, y) is 2 sq. units. \t\t\u2234 O lies on the bisector of \u2220BCD, i.e., AC. \t\t21 2 \u22124 4\u2212x =2 AB 4 \u22124 4\u2212y \t\t\u221202 4\u2212x =4\u21d2 2y \u2212 8 =4 4\u2212y \t\t\u21d2 |y \u2212 4| = 2 \u21d2 y \u2212 4 = \u00b1 2 \t\t \u21d2 y \u2212 6 = 0\u2003 or\u2003 y \u2212 2 = 0. \t37.\t The required locus is a circle of radius 4 units and passing through point (2, 4). \t\tLet P(x, y) be any point on the locus. \u2234 (x \u2212 2)2(y \u2212 4)2 = 4 x2 + y2 \u2212 4x \u2212 8y + 4 = 0. DC \t38.\t Given, P is the point of intersection of the diago- \t42.\t Given, any point on AD is equidistant from B nals of a square READ. and\u00a0C. \t\tIn a square, diagonals are equal and bisect to each \t\t\u2234 AD is the perpendicular bisector of BC. other.","15.16 Chapter 15 \t\tBy SAS congruence property, A \t\t \u2234 \u2220PBC = \u2220PCB = 30\u00b0 \t\t \u0394ADB \u2245 \u0394ADC. \u21d2 \u2220BPC = 180\u00b0 \u2212 (30\u00b0 + 30\u00b0) = 120\u00b0. \t\tBy CPCT, AB = AC. 4\t 5. \tLet A = (2, 6), B = (2, 8). 4\t 3.\t The required locus is the union B D C \t\tThe required locus is the perpendicular bisector of the locus passing through the of AB. origin and making angles of 45\u00b0 and 135\u00b0 with the X-axis in positive direction. \t\tLet K be the point of intersection of AB and its perpendicular bisector. \t\tThat is, x = |y| or y = |x|. \t44.\t Here, P is the point of intersection of bisectors of \t K = \uf8eb 2 + 2 , 8 + 6\uf8f6 = (2, 7). \b(1) \u2220B and \u2220C. \uf8ed\uf8ec 2 2 \uf8f7\uf8f8 A \t\tSlope of AB = 8\u22126 is not defined. P 2\u22122 BC \t\t\u2234 \u0007AB is parallel to Y-axis. The required line is parallel to X-axis\b(2) \t\t\u2234 T\u0007 he required line is y = 7 (From Eqs. (1) and (2)). Hints and Explanation","1162CChhaapptteerr TKriingeomnaotmicsetry REMEMBER Before beginning this chapter, you should be able to: \u2022 Know the di\ufb00erent types of angles and their values \u2022 Understand the definitions of triangles and their types KEY IDEAS After completing this chapter, you should be able to: \u2022 Know the systems of measurement of angle and their inter-relations \u2022 Understand trigonometric ratios and identities \u2022 Learn the trigonometric ratios of compound angles Figure 1.1","16.2 Chapter 16 INTRODUCTION The word trigonometry is originated from the Greek word \u2018tri\u2019 means three, \u2018gonia\u2019 means angle and \u2018metron\u2019 means measure. Hence, the word trigonometry means three angle measure, i.e., it is the study of geometrical figures which have three angles, i.e., triangles. The great Greek mathematician Hipapachus of 140 BCE gave relation between the angles and sides of a triangle. Further trigonometry is developed by Indian (Hindu) mathematicians. This was migrated to Europe via Arabs. Trigonometry plays an important role in the study of Astronomy, Surveying, Navigation and Engineering. Now a days it is used to predict stock market trends. ANGLE A measure formed between two rays having a common initial point is B called an angle. The two rays are called the arms or sides of the angle and the common initial point is called the vertex of the angle. \u2022 In the Fig. 16.1, OA is said to be the initial side and the other ray OB \u03b8 \u2022 is said to be the terminal side of the angle. C A The angle is taken positive when measured in anti-clockwise Figure 16.1 direction and is taken negative when measured in clockwise direction (see Fig. 16.2). BO Initial side A \u2022 (\u2013\u03b8) \u03b8 Terminal side C \u2022\u2003 B Figure 16.2 Systems of Measurement of Angle We have the following systems of the measurement of angle. Sexagesimal System In this system, the angle is measured in degrees (\u00b0). Degree\u2002 When the initial ray is rotated through \uf8eb 1\uf8f6 of one revolution, we say that an angle of \uf8ec\uf8ed 360\uf8f7\uf8f8 one degree (1\u00b0) is formed at the initial point. A degree is divided into 60 equal parts and each part is called one minute (1 m). Further, a minute is divided into 60 equal parts called seconds (\u2033). So, 1 right angle = 90\u00b0 1\u00b0 = 60\u2032 (minutes) and 1\u2032 = 60\u2033 (seconds) \u2002Note\u2002\u2002 This system is also called the British system. Centesimal system In this system, the angle is measured in grades.","Trigonometry 16.3 Grade\u2002 When the initial ray is rotated through \uf8eb 1\uf8f6 of one revolution, an angle of one grade is \uf8ed\uf8ec 400\uf8f7\uf8f8 said to be formed at the initial point. It is written as 1g. Further one grade is divided into 100 equal parts called minutes and one minute is further divided into 100 equal parts called seconds. So, 1 right angle = 100g 1g = 100\u2032 (minutes) and 1\u2032 = 100\u2033 (seconds) \u2002Note\u2002\u2002 This system is also called the French system. Circular System O rr In this system, the angle is measured in radians. Radian\u2002 The angle subtended by an arc of length equal to the radius of a 1c circle at its centre is said to have a measure of one radian (see Fig. 16.3). r It is written as 1c. Figure 16.3 \u2002Note\u2002\u2002 This measure is also known as radian measure. Relation Between the Units of the Three Systems When a rotating ray completes one revolution, the measure of angle formed about the vertex is 360\u00b0 or 400\u2009g or 2\u03c0\u2009c. So, 360\u00b0 = 400g = 2\u03c0 c (or ) 90\u00b0 = 100g = \u03c0c . 2 For convenience, the above relation can be written as, D = G = R . 90 100 \u03c0 2 Where, D denotes degrees, G grades and R radians. Remember 1.\t 1\u00b0 = \u03c0 radians = 0.0175 radians (approximately). 180 2.\t 1c = 180 degrees = 57\u00b017\u203244\u2032\u2032 (approximately). \u03c0 \u2002Notes\u2002 1.\t The measure of an angle is a real number. 2.\t If no unit of measurement is indicated for any angle, it is considered as radian measure.","16.4 Chapter 16 Example 16.1 Convert 60\u00b0 into circular measure. Solution Given, D = 60\u00b0 We have, D = R 90 \u03c0 2 So, 60 = R 90 \u03c0 2 2 \u00d7 \u03c0 = R 3 2 R = \u03c0 . 3 Hence, 60\u00b0 in circular measure is \u03c0. 3\u03c4 Example 16.2 Convert 180g into sexagesimal measure. Solution Given, G = 180g We have, D = G 90 100 So, D = 180 90 100 D = 9 \u00d7 90 = 162. 5 Hence, sexagesimal measure of 180g is 162\u00b0. Example 16.3 The angle measuring \u03c0 \u03c4when expressed in sexagesimal measure is ________. 4 Solution Given, R = \u03c0 . 4 We have, D = R . 90 \u03c0 2","Trigonometry 16.5 \u03c0 So, D = 4 90 \u03c0 2 D = 2 \u00d7 90 = 45\u00b0 4 Hence, the sexagesimal measure of \u03c0c \u03c4 is 45\u00b0. 4 TRIGONOMETRIC RATIOS B Let AOB be a right triangle with \u2220AOB as 90\u00b0. Let hypotenuse Opposite side to \u03b8 . \u2220OAB be \u03b8. Notice that 0\u00b0 < \u03b8 < 90\u00b0. That is, \u03b8 is an acute angle (see Fig. 16.4). We can define six possible ratios among the three \u03b8 sides of the triangle AOB, known as trigonometric AO ratios. They are defined as follows. Adjacent side to \u03b8 . 1.\t Sine of the angle \u03b8 or simply sin\u03b8: Figure 16.4 sin \u03b8 = Side opposite to angle \u03b8 = OB . Hypotenuse AB 2.\t Cosine of the angle \u03b8 or simply cos\u2009\u03b8: cos\u03b8 = Side adjacent to angle \u03b8 = OA . Hypotenuse AB 3.\t Tangent of the angle \u03b8 or simply tan\u2009\u03b8: tan \u03b8 = Side opposite to \u03b8 = OB . Side adjacent to \u03b8 OA 4.\t Cotangent of the angle \u03b8 or simply cot\u2009\u03b8: cot \u03b8 = Side adjacent to \u03b8 = OA . Side opposite to \u03b8 OB 5.\t Cosecant of the angle \u03b8 or simply cosec\u2009\u03b8: cosec \u03b8 = Hypotenuse = AB . Side opposite to \u03b8 OB 6.\t Secant of the angle \u03b8 or simply sec\u2009\u03b8: sec\u03b8 = Hypotenuse = AB . Side adjacent to \u03b8 OA Observe that, 1.\t cosec\u03b8 = 1 , sec\u03b8 = 1 and cot\u03b8 = 1 . sin \u03b8 cos\u03b8 tan \u03b8 2.\t tan \u03b8 = sin \u03b8 and cot\u03b8 = 1 = cos\u03b8 . cos\u03b8 tan \u03b8 sin \u03b8","16.6 Chapter 16 Example 16.4 R Q If cos\u03b8 = 3 , then find the values of tan\u2009\u03b8, cosec\u2009\u03b8. 5 Solution Given, cos\u03b8 = 3 5 \u03b8 Let PQR be the right triangle such that \u2220QPR = \u03b8 (see Fig. 16.5) P Figure 16.5 Assume that PQ = 3 and PR = 5. Then, QR = PR2 \u2212 PQ2 = 25 \u2212 9 = 4. So, tan \u03b8 = Opposite side = QR = 4 and cosec\u03b8 = Hypotenuse = PR = 5 . Adjacent side PQ 3 Opposite side QR 4 Pythagorean Triplets 2.\u2002 5, 12, 13 1.\t 3, 4, 5\t\t\t 4.\u2002 7, 24, 25 3.\t 8, 15, 17\t\t\t 5.\t 9, 40, 41 Trigonometric Identities 1.\t sin2\u2009\u03b8 + cos2\u2009\u03b8 = 1 2.\t sec2\u2009\u03b8 \u2212 tan2\u2009\u03b8 = 1 3.\t cosec2\u2009\u03b8 \u2212 cot2\u2009\u03b8 = 1 Values of Trigonometric Ratios for Specific Angles Trigonometric Ratios 0\u00b0 30\u00b0 Angle 60\u00b0 90\u00b0 0 45\u00b0 3 1 sin\u2009\u03b8 1 2 0 2 1 1 \u221e 2 2 1 \u221e cos\u2009\u03b8 1 31 3 0 22 2 tan\u2009\u03b8 0 1 1 3 3 2 cosec\u2009\u03b8 \u221e2 2 1 sec\u2009\u03b8 12 3 2 3 cot\u2009\u03b8 \u221e 3 1","Trigonometry 16.7 From the above table, we observe that 1.\t sin\u2009\u03b8 = cos\u2009\u03b8, tan\u2009\u03b8 = cot\u2009\u03b8 and sec\u2009\u03b8 = cosec\u2009\u03b8, if \u03b8 = 45\u00b0. 2.\t sin\u2009\u03b8 and tan\u2009\u03b8 are increasing functions in 0\u00b0 \u2264 \u03b8 \u2264 90\u00b0. 3.\t cos\u2009\u03b8 is a decreasing function in 0\u00b0 \u2264 \u03b8 \u2264 90\u00b0. Example 16.5 Find the value of sin\u200960\u00b0 + 2tan\u200945\u00b0 \u2212 cos\u200930\u00b0. Solution sin\u200960\u00b0 + 2tan\u200945\u00b0 \u2212 cos\u200930\u00b0 = 3 + 2(1) \u2212 3 = 2 2 2 \\\\ sin\u200960\u00b0 + 2tan\u200945\u00b0 \u2212 cos\u200930\u00b0 = 2. Example 16.6 Using the trigonometric table and evaluate (a)\t sin245\u00b0 + cos245\u00b0 (b)\t sec230\u00b0 \u2212 tan230\u00b0 Solution (a)\t sin245\u00b0 + cos245\u00b0 = \uf8eb 1 \uf8f62 + \uf8eb 1 \uf8f62 \uf8ed\uf8ec 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \uf8f8\uf8f7 \t 1 1 = 2 + 2 = 1. Hence, sin245\u00b0 + cos245\u00b0 = 1. (b)\t sec230\u00b0 \u2212 tan230\u00b0 = \uf8eb 2 \uf8f62 \u2212 \uf8eb 1 \uf8f62 \uf8ec\uf8ed 3 \uf8f8\uf8f7 \uf8ed\uf8ec 3 \uf8f8\uf8f7 \t 4 1 3 = 3 \u2212 3 = 3 = 1. Hence, sec230\u00b0 \u2212 tan230\u00b0 = 1. Example 16.7 Find the values of tan 30\u00b0 + tan 60\u00b0 and tan 90\u00b0; what do you observe? 1 \u2212 tan 30\u00b0 tan 60\u00b0 Solution tan 30\u00b0 + tan 60\u00b0 1 \u2212 tan 30\u00b0 tan 60\u00b0","16.8 Chapter 16 1 1+3 = 3+ 3 = 3 = 4 1 1\u22121 0 1\u2212 3 \u00d7 3 = not defined or \u221e. = tan\u200990\u00b0 Hence, tan 30\u00b0 + tan 60\u00b0 = tan 90\u00b0. 1 \u2212 tan 30\u00b0 tan 60\u00b0 Trigonometric Ratios of Compound Angles 1.\t sin\u2009(A + B) = sin A cos B + cos A sin B and \t sin\u2009(A \u2212 B) = sin A cos B \u2013 cos A sin B. 2.\t cos\u2009(A + B) = cos A cos B \u2013 sin A sin B and \t cos\u2009(A \u2212 B) = cos A cos B + sin A sin B. 3.\t tan ( A + B) = tan A + tan B and 1 \u2212 tan A tan B \t tan ( A \u2212 B ) = tan A \u2212 tan B . 1 + tan A tan B Also, by taking A = B in the above relations, we get, 1.\t sin 2A = 2sin A\u2009cos A. 2.\t cos 2A = cos2A \u2212 sin2A = 2 cos2A \u2212 1 = 1 \u2013 2 sin2A. 3.\t tan 2A = 1 2 tan A . \u2212 tan2 A Example 16.8 Find the value of sin15\u00b0. Solution We have, sin15\u00b0 = sin\u2009(45\u00b0 \u2212 30\u00b0) = sin 45\u00b0 cos 30\u00b0 \u2212 cos 45\u00b0 sin 30\u00b0 = 1 \u00d7 3 \u2212 1 \u00d7 1 = 3 \u22121 2 2 2 2 22 \\\\\u2009sin15\u00b0 = 3 \u22121. 22 Example 16.9 Find the value of tan 75\u00b0. Solution We have, tan 75\u00b0 = tan\u2009(45\u00b0 + 30\u00b0)","Trigonometry 16.9 tan 45\u00b0 + tan 30\u00b0 1 \u2212 tan 45\u00b0 \u22c5 tan 30\u00b0 1+ 1 3 +1. 3 3 \u22121 = = 1 1\u22121\u00d7 3 \uf8eb 3 +1\uf8f6\uf8eb 3 + 1\uf8f6 = 3 + 1+ 2 3 =2+ 3 = \uf8ec\uf8ed\uf8ec 3 \u2212 1 \uf8f7\uf8f8\uf8f7 \uf8ec\uf8ed\uf8ec 3 + 1 \uf8f7\uf8f7\uf8f8 2 \u2234 tan 75\u00b0 = 3 + 1 or 2 + 3. 3 \u2212 1 Example 16.10 Eliminate \u03b8 from the equations a = x sec\u2009\u03b8 and b = y tan\u2009\u03b8. Solution We know that trigonometric ratios are meaningful when they are associated with some \u03b8, i.e., we cannot imagine any trigonometric ratio without \u03b8. Eliminate \u03b8 means, eliminating the trigonometric ratios by using suitable identity. Given, a = x sec\u2009\u03b8 and b = y tan\u2009\u03b8 a = sec\u03b8 and b = tan \u03b8 . x y We know that, sec2\u2009\u03b8 \u2212 tan2\u2009\u03b8 = 1. So,\t \uf8eb a\uf8f62 \u2212 \uf8eb b\uf8f62 = 1 \uf8ed\uf8ec x \uf8f8\uf8f7 \uf8ec\uf8ed y \uf8f7\uf8f8 a2 \u2212 b2 = 1. x2 y2 Hence, the required equation is a2 \u2212 b2 = 1. x2 y2 Example 16.11 Find the relation obtained by eliminating \u03b8 from the equations x = a cos\u2009\u03b8 + b sin\u2009\u03b8 and y = a sin\u2009\u03b8 \u2013 b cos\u2009\u03b8. Solution Given, x = a cos\u2009\u03b8 + b sin\u2009\u03b8 \tx2 = (a cos\u2009\u03b8 + b sin\u2009\u03b8)2 \t = a2cos2 \u03b8 + b2sin2 \u03b8 + 2ab cos\u2009\u03b8 sin\u2009\u03b8.","16.10 Chapter 16 Also y = a sin\u2009\u03b8 \u2212 b cos\u2009\u03b8 \ty2 = a2sin2\u2009\u03b8 + b2cos2\u2009\u03b8 \u2212 2ab sin\u2009\u03b8 cos\u2009\u03b8 \tx2 + y2 = a2 (sin2\u2009\u03b8 + cos2\u2009\u03b8) + b2 (cos2\u2009\u03b8 + sin2\u2009\u03b8) \t = a2 + b2. Hence, the required relation is x2 + y2 = a2 + b2. Example 16.12 Eliminate \u03b8 from the equations P = a cosec\u2009\u03b8 and Q = a cot\u2009\u03b8. Solution Given, P = a cosec\u2009\u03b8 and Q = a cot\u2009\u03b8 P = cosec\u03b8 and Q = cot\u03b8 . a a We know that, cosec2\u2009\u03b8 \u2212 cot2\u2009\u03b8 = 1 \uf8ebP \uf8f62 \u2212 \uf8ebQ \uf8f62 =1 \uf8ed\uf8ec a \uf8f8\uf8f7 \uf8ec\uf8ed a \uf8f8\uf8f7 P\u20092 \u2212 Q2 = a2. Hence, the required relation is P\u20092 \u2212 Q2 = a2. Example 16.13 Eliminate \u03b8 from the equations s = sin\u2009\u03b8 + cosec\u2009\u03b8 and r = sin\u2009\u03b8 \u2013 cosec\u2009\u03b8. Solution \tGiven, s = sin\u2009\u03b8 + cosec\u2009\u03b8\b (1) \tr = sin\u2009\u03b8 \u2013 cosec\u2009\u03b8\b(2) Adding Eqs. (1) and (2), we get s + r = 2 sin\u2009\u03b8 \t sin \u03b8 = s + r . \b(3) get 2 Subtracting Eq. (2) from Eq. (1), we s \u2212 r = 2 cosec\u2009\u03b8 \t cosec\u03b8 = s \u2212 r . \b(4) 2 Multiplying Eqs. (3) and (4), we get sin \u03b8 \u22c5 cosec\u03b8 = \uf8eb s + r \uf8f6\uf8eb s \u2212r \uf8f6 \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \uf8f8\uf8f7 sin \u03b8 \u22c5 1 = s2 \u2212 r2 sin \u03b8 4 1 s2 \u2212 r 2 (or ) s2 \u2212 r 2 = 4. 4 = Hence, by eliminating \u03b8, we obtain the relation s2 \u2212 r2 = 4.","Trigonometry 16.11 Example 16.14 If sin (A + B) = 3 and cosec A = 2, then find A and B. 2 Solution Given, sin (A + B) = 3 2 \t sin\u2009(A + B) = sin 60\u00b0 \tA + B = 60\u00b0\b(1) \tcosec A = 2 = cosec 30\u00b0 \tA = 30\u00b0\b(2) From Eqs. (1) and (2), we have A = 30\u00b0 and B = 30\u00b0. Example 16.15 Find the length of the chord which subtends an angle of 90\u00b0 at the centre \u2018O\u2019 and which is at a distance of 6 cm from the centre. Solution O Let the chord be PQ and OR be the distance of chord from the centre of circle (see Fig. 16.6). Given \u2220POQ = 90\u00b0 and OR = 6 cm. Clearly, \u0394POR \u2245 \u0394QOR, by SSS axiom of congruency. \u2220POR = \u2220QOR = 1 \u2220POQ = 45\u00b0. 45\u00b0 45\u00b0 2 P RQ In \u2206POR, tan 45\u00b0 = PR . Figure 16.6 OR 1= PR 6 \tPR = 6 cm. \\\\ The length of the chord PQ = 2 PR = 12 cm. Example 16.16 Evaluate 1 \u2212 cos\u03b8 . 1 + cos\u03b8 Solution Given 1 \u2212 cos\u03b8 1 + cos\u03b8","16.12 Chapter 16 Rationalize the denominator, i.e., 1 \u2212 cos\u03b8 . 1 \u2212 cos\u03b8 1 + cos\u03b8 1 \u2212 cos\u03b8 = (1 \u2212 cos\u03b8 )2 1 + cos2 \u03b8 = (1 \u2212 cos\u03b8 )2 (\u2235 sin2 \u03b8 + cos2 \u03b8 = 1) sin2 \u03b8 = 1 \u2212 cos \u03b8 sin \u03b8 = 1 \u2212 cos\u03b8 sin \u03b8 sin \u03b8 = cosec\u03b8 \u2212 cot\u03b8 . \u2234 1 \u2212 cos\u03b8 = cosec\u03b8 \u2212 cot\u03b8. 1 + cos\u03b8 Example 16.17 If sin \u03b1 and cos \u03b1 are the roots of the equation ax2 \u2212 bx \u2212 1 = 0, then find the relation between a and b. Solution The given equation is ax2 \u2212 bx \u2212 1 = 0. Here, a = a, b = \u2212b and c = \u22121. sin \u03b1 + cos \u03b1 = \u2212b = \u2212(\u2212b) = b a a a sin \u03b1 \u22c5 cos \u03b1 = c = \u22121 . a a Consider, sin \u03b1 + cos \u03b1 = \u2212b a Squaring on both sides, (sin \u03b1 + cos\u03b1 )2 = \uf8eb \u2212b \uf8f62 \uf8ec\uf8ed a \uf8f7\uf8f8 = 1 + 2 \uf8eb \u22121 \uf8f6 = b2 \uf8ec\uf8ed a \uf8f8\uf8f7 a2 =1\u2212 b2 = 2 a2 a \u2234 a2 \u2212 b2 = 2a.","Trigonometry 16.13 Example 16.18 If cos \u03b1 = 2 and sin \u03b2 = 1 , then find cos(\u03b1 \u2212 \u03b2).\u2009\u2009\u2009\u2009 3 4 Solution Given, cos \u03b1 = 2 \u21d2 sin \u03b1 = 5 3 3 sin \u03b2 = 1 \u21d2 cos \u03b2 = 15 4 4 cos(\u03b1 \u2212 \u03b2) \u2013 cos \u03b1 \u00b7 cos \u03b2 + sin \u03b1 \u00b7 sin \u03b2 = 2 \u00d7 15 + 5 \u00d7 1 3 4 3 4 cos(\u03b1 \u2212 \u03b2 ) = 2 15 + 5. 12 Example 16.19 Express the following as a single trigonometric ratio: (a)\t 3 cos\u03b8 \u2212 sin\u03b8 (b)\t sin\u2009\u03b8 \u2013 cos\u2009\u03b8 Solution 3 cos\u03b8 \u2212 sin\u03b8 (a)\t Given, = 2 \uf8eb 3 cos\u03b8 \u2212 1 sin \u03b8 \uf8f6 (b)\t Here, \uf8ec\uf8ed\uf8ec 2 2 \uf8f7\uf8f7\uf8f8 = 2(cos\u03b8 \u22c5 cos 30\u00b0 \u2212 sin\u03b8 \u22c5 sin 30\u00b0) = 2(cos (\u03b8 + 30\u00b0)) \u21d2 3 cos\u03b8 \u2212 sin\u03b8 = 2 cos(\u03b8 + 30\u00b0). sin\u2009\u03b8 \u2212 cos\u2009\u03b8 = 2 \uf8eb sin \u03b8 \u2212 cos\u03b8 \uf8f6 \uf8ed\uf8ec 2 \uf8f7\uf8f8 = 2 \uf8eb 1 sin\u03b8 \u2212 1 cos\u03b8 \uf8f6 \uf8ec\uf8ed 2 2 \uf8f7\uf8f8 = 2 \uf8f0\uf8ee\uf8efsin \u03b8 cos \uf8eb \u03c0 \uf8f6 \u2212 cos\u03b8 sin \uf8eb \u03c0 \uf8f6\uf8f9 \uf8ec\uf8ed 4 \uf8f7\uf8f8 \uf8ec\uf8ed 4 \uf8f7\uf8f8\uf8fa\uf8fb = 2 sin \uf8ed\uf8eb\uf8ec\u03b8 \u2212 \u03c0 \uf8f7\uf8f6\uf8f8 . (\u2235 sin( A \u2212 B ) = sin A cos B \u2212 cos A sin B) 4","16.14 Chapter 16 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t The value of 144\u00b0 in circular measure = ______. 1\t 7.\t tan(A + B) = 3 and sin A = 1 , then the value 2 of B in radians is ______. \t2.\t The value of sin 30\u00b0 \u22c5 sin 45\u00b0 \u22c5 cosec 45\u00b0 \u22c5 cos 30\u00b0 = ______. \t18.\t In \u0394ABC, the lengths of the three sides AB, BC and CA are 28 cm, 96 cm and 100 cm respectively. \t3.\t If 1 \u2212 tan2 60\u00b0 = cos X, then the value of X Find the value of cos C. is __1_+__t_an. 2 60 1\t 9.\t If sin A = cos B, where A and B are acute angles, \t4.\t The value of log [(sec\u2009\u03b8 + tan\u2009\u03b8) (sec\u2009\u03b8 \u2013 tan\u2009\u03b8)] is then A + B = ______. ______. \t20.\t ABC is a right isosceles triangle, right angled at B. \t5.\t If cos 13\u00b0 + sin 13\u00b0 = tan A, then A = ______. Then sin2A + cos2C = ______. cos 13\u00b0 \u2212 sin 13\u00b0 2\t 1.\t Evaluate: sin260\u00b0 cos245\u00b0 cos260\u00b0 cosec290\u00b0. \t6.\t If sin \u03b8 = 1 and 0\u00b0 < \u03b8 < 90\u00b0, then cos 2\u03b8 \t22.\t If sin \u03b8 = 3 and \u03b8 is acute, then find the value of 2 5 = ______. tan\u03b8 \u2212 2 cos\u03b8 \t7.\t If cosec\u2009\u03b8 \u2013 cot\u2009\u03b8 = x, then cosec\u2009\u03b8 + cot\u2009\u03b8 = 3 sin\u03b8 + sec\u03b8 . ______. \t8.\t The value of sin 20\u00b0 cos 70\u00b0 + cos 20\u00b0 sin 70\u00b0 is \t23.\t Find the values of the cos 15\u00b0. sin 23\u00b0 coses 23\u00b0 + cos 23\u00b0 sec 23\u00b0 2\t 4.\t Evaluate: cosec2 30\u00b0 + sec2 60\u00b0 + tan2 30\u00b0. ______. 2\t 5.\t Convert \u03c0c into the other two systems. 15 \u03c4 PRACTICE QUESTIONS \t9.\t If A + B = 45\u00b0, then (1 + tan A)(1 + tan B) = 2\t 6.\t In the adjoining figure, find the values of tan B. ______. 1\t 0.\t If tan \u03b8 = 5 and tan \u03c6 = 1 , then \u03b8 + \u03d5 = ______. A 6 11 \t11.\t If A \u2212 B = 45\u00b0 and tan A \u2212 tan B = 3, then tan\u2009A 5 \u22c5 tan\u2009B = ______. 1\t 2.\t If cos A + sin A = 2(cos 1 sin A) ; (0\u00b0 < A< A\u2212 90\u00b0), then sin2A = ______. B 3D 2C \t13.\t The value of (sin A \u2013 cos A)2 + (sin A + cos A)2 is \t27.\t Simplify and express sec4\u03b1 \u2212 tan4\u03b1 in their least ______. exponents. \t14.\t If A + B = 60\u00b0, then the value of sin\u2009A cos\u2009B Convert\u2009\uf8ec\uf8ed\uf8eb 200 g + cos\u2009A sin\u2009B = ______. 3 1\t 5.\t In \u0394ABC, the value of cos (A \u2212 B) \u2212 cos (C) is \t28.\t \uf8f6 \u2009into other two systems. ______. \uf8f7\uf8f8 \t16.\t ABC is right triangle, right angled at A, then 2\t 9.\t Convert 270\u00b0 into other two systems. tan\u2009B \u22c5 tan\u2009C = ______. \t30.\t Evaluate: cos 0\u00b0 + 2 sec 45\u00b0 \u2212 3 tan 30\u00b0.","Trigonometry 16.15 Short Answer Type Questions \t31.\tIn the adjoining figure, find the value of sin C. 3\t 7.\t If cos \u03b1 = 12 and sin \u03b2 = 4 , then find sin(\u03b1 + \u03b2). 13 5 A \t38.\t Express sin\u2009\u03b8 in terms of cot\u2009\u03b8. 5 \t39.\t If sin \u03b1 = 4 , then find the value of sin 2\u03b1. B 3D 2 C 5 \t32.\t A wheel makes 200 revolutions in 2 minutes. Find \t40.\t Find the value of sin 2\u03b1, if sin \u03b1 + cos\u03b1 = 1 . the measure of the angle it describes at the centre 3 in 24 seconds. \t41.\t The length of the minutes hand of a wall clock 3\t 3.\t Find the length of the chord subtending an angle is 36 cm. Find the distance covered by its tip in of 120\u00b0 at the centre of the circle whose radius is 4 cm. 35 minutes. 3\t 4.\t Find the value of tan 75\u00b0. 4\t 2.\t If tan 2\u03b1 = 3 , then find tan\u2009\u03b1. 3\t 5.\t If sec2\u03b1 + cos2\u03b1 = 2, then find the value of sec\u2009\u03b1 + 4 cos\u2009\u03b1. \t43.\t If A + B = 45\u00b0, then find the value of tan\u2009A + tan B 3\t 6.\t Eliminate \u03b8 from the equations, a = x sin\u2009\u03b8 \u2013 y + tan\u2009A tan\u2009B. cos\u2009\u03b8 and b = x cos\u2009\u03b8 + y sin\u2009\u03b8. \t44.\t If sin(A + B) = 3 and cot (A \u2212 B) = 1, then find 2 A and B. \t45.\t If sin\u2009\u03b8 and cos\u2009\u03b8 are the roots of the quadratic equation lx2 \u2212 mx \u2212 n = 0, then find the relation between l, m and n. Essay Type Questions \t46.\tShow that 3(sin x + cos x)4 \u2212 6(sin x + cos x)2 + 4\t 9.\t One of the angles of a rhombus is 60\u00b0 and the PRACTICE QUESTIONS length of the diagonal opposite to it is 6 cm. Find 4(sin\u20096x + cos\u20096x) = 1. the area of the rhombus (in sq.cm). \t47.\t If cos2\u03b1 + cos \u03b1 = 1, then find the value of 4 sin2\u03b1 5\t 0.\t If \u03b1 + \u03b2 + \u03b3 = 90\u00b0, then tan\u2009\u03b1 + tan\u2009\u03b2 + tan\u2009\u03b1 tan \u03b2\u2009 + 4 sin4\u03b1 + 2. cot\u2009\u03b3 is ______. 4\t 8.\t Obtain the relation by eliminating \u03b8 from the equations, x = a + r cos\u2009\u03b8 and y = b + r sin\u2009\u03b8. CONCEPT APPLICATION Level 1 \t1.\t The length of the minutes hand of a wall clock is \t\t(a) 1\t\t (b) 1 6 cm. Find the distance covered by the tip of the 3 minutes hand in 25 minutes. \t\t(c) 0\t\t (d) 8 \t\t(a) 270 cm \t\t (b) 110 cm \t3.\t The value of sin\u2009\u03b8 in terms of tan\u2009\u03b8 is ______. 7 \t\t(a) tan\u03b8 \t(b) tan2 \u03b8 \t\t(c) 88 cm \t\t (d) 110 cm 1 \u2212 tan2 \u03b8 1 + tan2 \u03b8 7 7 \t2.\t The value of tan 10\u00b0 \u00b7 tan 20\u00b0 \u00b7 tan 45\u00b0 \u00b7 tan 70\u00b0 \u00b7 \t\t(c) tan2 \u03b8 \t(d) tan\u03b8 tan 80\u00b0 = 1 \u2212 tan2 \u03b8 1 + tan2 \u03b8","16.16 Chapter 16 \t4.\t The value of sin260\u00b0 + cos230\u00b0 \u2212 sin245\u00b0 is \t\t(a) \u03c0 \u03c4 \t\t (b) \u03c0 \u03c4 _______. 6 3 \t\t(a) 1\t\t (b) sin 90\u00b0 \t\t(c) 5\u03c0 \u03c4 \t\t (d) 7\u03c0 \u03c4 (d) Both (a) and (b) 6 6 \t\t(c) 1 \t\t 2 3\u03c0 c \t11.\t The value of 5\u03c4 in sexagesimal measure is \uf8ec\uf8ed\uf8eb1 1 \uf8f6 ______. \t5.\t The simplified value of cosec2\u03b1 + sec \u03b1 \uf8f7\uf8f8 1 \t\t(a) 216\u00b0\t\t (b) 144\u00b0 sec \u03b1 \uf8ed\uf8ec\uf8eb1 \u2212 \uf8f6 is ________. \t\t(c) 128\u00b0\t\t (d) 108\u00b0 \uf8f8\uf8f7 1\t 2.\t The value of logsec\u2009\u03b8 (1 \u2212 sin2\u2009\u03b8) is ______. \t\t(a) 1\t\t (b) 0 \t\t(c) 2\t\t (d) \u22121 \t\t(a) 2\t (b) \u22122\t (c) 0\t (d) 1 \t6.\t A wheel makes 240 revolutions in one minute. \t13.\t If sin A = 3 and A is an acute angle, then find Find the measure of the angle it describes at the 2 centre in 15 seconds. the value of tan A \u2212 cot A . \t\t(a) 60 \u03c0\u2009c\t\t (b) 120 \u03c0\u2009c 3 + cosecA \t\t(c) 8 \u03c0\u2009c\t\t (d) None of these \t\t(a) \u22122 \t\t (b) 2 5 5 \t7.\t If 3 tan\u2009A = 4, then find the value of \t\t(c) 2 3\t (d) \u22122 3+2 2 sin A \u2212 7 cos A . 3 cos A + 4 1\t 4.\t The length of an arc, which subtends an angle \t\t(a)\t \u221213 \t\t (b) 13 of 30\u00b0 at the centre of the circle of radius 42 cm 29 11 is ______. \t\t(c) \u221e\t\t (d) 29 \t\t(a) 22 cm\t\t (b) 44 cm 13 \t\t(c) 11 cm\t\t (d) 22 cm 7 PRACTICE QUESTIONS 1 \t8.\t If tan\u2009\u03b1 = 3 and tan \u03b2 = 2 , then which of the 1\t 5.\t If sin4 \u03b8 + cos4 \u03b8 = 1 , then find sin\u2009\u03b8cos\u2009\u03b8. 2 following is true? \t\t(a) \u03b1 + \u03b2 = \u03c0c \t(b) \u03b1\u03b2 = \u03c0c \t\t(a) \u00b1 1 \t\t (b) \u00b1 1 4 4 8 4 \u03c0c \t\t(c) \u00b11\t\t (d) \u00b1 1 4 2 \t\t(c) \u03b1 \u2212 \u03b2 = \t (d) None of these a b \t16.\t If sin \u03b8 = , then cos\u2009\u03b8 and tan\u2009\u03b8 in terms of a and \t9.\t If tan A = 1 and tan B = 1 , then which of the b are 2 3 following is true? \t\t(a) b2 \u2212 a2 and b b b2 \u2212 a2 \u03c0c \u03c0c \t\t(a) A+B = 4 \t(b) A\u2212B = 4 \t\t(b) b and a \u03c0c \u03c0c b2 \u2212 a2 b2 \u2212 a2 4 4 \t\t(c) 2A + B = \t(d) A + 2 = a2 \u2212 b2 b a a2 \u2212 b2 \t\t(c) and \t\t(d) a \t10.\t If sin A = 1 and 90\u00b0 < A < 180\u00b0, then the value b2 \u2212 a2 b2 \u2212 a2 2 b and of A in circular measure is ______.","Trigonometry 16.17 \t17.\t The distance covered by the tip of a minute hand 2\t 4.\t FindthemeasureofangleA,if(2sin\u2009A+1)(2sin\u2009A\u22121) in 35 minutes is 33 cm. What is the length of the = 0. minute hand? \t\t(a) 75\u00b0\t\t (b) 60\u00b0 \t\t(a) 6 cm\t\t (b) 9 cm \t\t(c) 45\u00b0\t\t (d) 30\u00b0 \t\t(c) 10 cm\t\t (d) 12 cm 2\t 5.\t The value of cos\u20092\u03b8 in terms of cot\u2009\u03b8 is ______. 4 \t18.\t If sin \u03b1 = 5 , where (0\u00b0 \u2264 \u03b1 \u2264 90\u00b0), then find cot2 \u03b8 +1 cot2 \u03b8 \u22121 sin 2\u03b1. \t\t(a) \t\t(a) 12 \t\t (b) \u2212 24 \t\t(b) \u2212 1 + cot 2\u03b8 25 25 cot2 \u03b8 \u22121 \t\t(c) 25 \t\t (d) 24 \t\t(c) cot2 \u03b8 \u22121 24 25 cot2 \u03b8 +1 \t19.\t In a \u0394ABC, cos \uf8eb A + B\uf8f6 = _____ . \t\t(d) 1 \u2212 cot2 \u03b8 \uf8ec\uf8ed 2 \uf8f8\uf8f7 1 + cot2 \u03b8 \t\t(a) cos C \t\t (b) \u2212 sin C \t26.\t The simplified value of sin4 \u03b1 + cos4 \u03b1 + 1 sin2 2\u03b1 2 2 2 cos \uf8eb A \u2212 B \uf8f6 sin C is \uf8ed\uf8ec 2 \uf8f7\uf8f8 2 \t\t(c) \t(d) \t\t(a) \u22121\t\t (b) sin\u2009\u03b1 + cos\u2009\u03b1 \t20.\t sin4\u03b8 \u2212 cos4\u03b8 = \t\t (c) 0\t\t (d) 1 \t\t(a) \u22121\t\t (b) cos\u20092\u03b8 (1 + sin 2\u03b8 ) \uf8f0\uf8ef\uf8eewhere \uf8ef\uf8ee\uf8f00, \u03c0 \uf8f9\uf8f9 1\u2212 cos2 \u03b8 4 \uf8fb\uf8fa \uf8fa\uf8fb \t\t (c) 2sin2\u03b8 \u2212 1\t (d) sin\u20092\u03b8 2\t 7.\t \u03b8 \u2208 = \t21.\t If P : Q = tan\u20092A : cos\u2009A and Q : R = cos\u20092A : \t\t(a) cosec2\u03b8\t\t (b) 1 sin\u20092A, then P : R is _________. \t\t(c) 1 + cot\u2009\u03b8\t (d) 1 + tan\u2009\u03b8 \t\t(a) tan\u20092A\t\t (b) 2sin\u2009A PRACTICE QUESTIONS \t28.\t If A and B are complementary angles, then the \t\t(c) 1\t\t (d) sec\u2009A sin2 A + sin2 B \t22.\t If A = sin\u2009\u03b8 + cos\u2009\u03b8 and B = sin\u2009\u03b8 \u2212 cos\u2009\u03b8, then value of cosec2A \u2212 tan2 B is _______. which of the following is true? \t\t(a) 0\t\t (b) 1 \t\t(a) A2 + B2 = 1 \t\t(c) \u22121\t\t (d) 2 \t\t(b) A2 \u2212 B2 = 2 \t\t(c) A2 + B2 = 2 2\t 9.\t Find the value of 4(sin430\u00b0 + cos430\u00b0) \u2212 3(cos245\u00b0 \t\t(d) 2A2 + B2 = 4 + sin290\u00b0). 1 \t\t(a) \u2212 1 \t\t (b) \u22122 2 2 2\t 3.\t If cot(A \u2212 B) = 1 and cos(A + B) = , then find B. \t\t(c) 2\t\t (d) 1 2 1\u00b0 \t\t(a) 42 2 \u22c5 4 3 \t30.\t If sec\u03b8 + tan\u03b8 = , then sec\u2009\u03b8 tan\u2009\u03b8 = \t\t(b) 7 1\u00b0 \u22c5 175 25 2 24 576 \t\t(a) \t\t (b) \t\t(c) 15 1\u00b0 \u22c5 27 175 2 576 576 \t\t(c) \t\t (d) \t\t(d) 60\u00b0"]
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