Exam Corner - STRAIGHT LINES

Name ................................................ OBJECTIVE EXAM - MATHEMATICS Batch.................... Roll No. ............... 24-10-2021 Straight Line Batch: LT-23 Normal G-1 23V/TP/M Section 1 - Only one correct option type (4, –1) 1. Find the equation of the perpendicular bisector of the line segment joining the points A(2,3) and B(6,–5) 1) x  2y  6  0 2) x  2y  6  0 3) 2x  2y  6  0 4) x  2y  4  0 2. The points (-2,-5) , (2,-2) (8,a) are collinear then the value of a is 1) 5 2) 5 3) 3 4) 10 2 2 2 2 3. The orthocentre of a triangle formed by (8,0) and (4,6) with the origin is 1)  4, 8  2) 4,3 3)  8 , 4  4) 3, 4  3   3  4. The three vertices of a parallelogram are (-1, 2) (5, 1) and (6, 5). The coordinates of fourth vertex is: 1) (6, 0) 2) (6, 5) 3) (0, 6) 4) (-6, 0) 5. If orthocentre and circum centre of triangle are respectively (1,1) and (3,2) , then the co ordinates of its centroid are :  1) 7 3 , 53  2) 53 , 7 3 3) (7,5) 4) (5,7) 6. If a line is drawn through the origin and parallel to the line x – 2y + 5 = 0 its equation is 1) x – 2y – 5 = 0 2) 2x+y = 0 3) x + 2y = 0 4) x – 2y = 0 7. If p be the length of perpendicular from origin on the line whose intercepts on the axes are a and b then: 1) p2 = a2 + b2 1 11 111 11 1 2) p2  a2  b2 3) p2  a2  b2 4) p  a2  b2 8. The mediansAD and BE of the triangle with verticesA(0, b), B (0, 0), C(a, 0) are mutually perpendicular if: 1) a = 2b 2) a = b/2 3) b = 2a 4) b = -2a 9. The equation of the line perpendicular to 3x + 2y = 8 and passes through midpoint of the line joining (5, -2) and (2, 2) : 1) 2x - 3y = 7 2) 2x + 3y = 7 3) 3x - 2y = 7 4) 3x + 2y = 7 10. The foot of the perpendicular from point (2,4) upon x + y = 1 is 1)  1 , 3  2)   1 , 3  3)  4 , 1 4)  3 , 1   2 2   2 2   3 2   4 2  11. The point on the axis of y which is equidistant from (–1,2) and (3,4) is 1) (0,4) 2) (0,5) 3) (5,0) 4) (1,2) 12. The value of a so that the points (3,5), (2, -3) and (4, a) are collinear is: 1) a = 13 2) a = 10 3) a = 32 4) a = 9 13. The equation of the line that has y-intercept 4 and is parallel to the line 2x – 3y = 7 1) 2x – 3y + 1 = 0 2) 2x – 3y + 12 = 0 3) x + y + 4 = 0 4) 2x + y + 6 = 0

23V/M 2 J - MATHEMATICS 14. The points A (0, –1), B (2, 1), C(0, 3), D(–2, 1) are the vertices of a 1) Square 2) Rectangle 3) Parallelogram 4) Rhombus 15. The equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, –1). Length of its side is 1) 1 2) 3 3) 2 4) 2 2 2 3 16. If the line x  y  1passes through the points (2, –3) and (4, –5), then (a, b) is: a b 1) (1, 1) 2) ( –1, 1) 3) (1, –1) 4) (–1, –1) 17. The reflection of the point (4, –13) in the line 5x+y+6 =0 is: 1) (–1, –14) 2) (3, 4) 3) (0, 0) 4) (1, 2) 18. Find the equation of the perpendicular bisector of the line segment joining the points A(2,3) and B(6,–5) 1) x  2y  6  0 2) x  2y  6  0 3) 2x  2y  6  0 4) x  2y  4  0 19. Find the equation of the line passing through the point (2,2) and catting off intercepts on the axes whose sum is 9 1) x  2y  6  0 2) 2x  y  6  0 3) 3x  y  6  0 4) x  y  6  0 20. If a + b +c =0, the straight line 2ax + 3by + 4c =0 passes through the fixed point  1) 43 , 43 2) (2,2)  3) 2, 43 4) (1,2) Section II - Numerical answer type The answer to each question is a NUMERICAL VALUE. Each question carries +4 marks for correct answer and 0 mark for wrong answer. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, –.30, 30.27, –127.30, 0.60) 21. The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is k then 10k = 22. The inclination of the line x–y+3=0 with the positive direction of x–axis in degree measure is 23. The angle between the lines 2x – y + 3 = 0 and x + 2y + 3 = 0 in degree measure is 24. The points (-2,-5) , (2,-2) (8,a) are collinear then the value of 2a is 25. A line is drawn through the points (3,4) and (5,6). If the line is extended to a point whose ordinate is –1, then the abscissa of that point is k then -k is

Name ................................................ OBJECTIVE EXAM - MATHEMATICS Batch.................... Roll No. ............... Straight Line Batch: LT-23 Normal G-1 24-10-2021 23V/TP/M 1. 1 2. 2 3. 1 4. 3 GFH IKJLet D be(x, y); Midpoint of AC = (5/2, 7/2) = x  5 , y1 5. 1 2 2 6. 4 Circum centre, centroid and orthocentre are collinear. SG:GO = 1:2 7. 2 Any line parallel to given line is x  2y    0 . The pts (0,0) lies on this line if 0  0    0 if   0 8. 1 is x – 2y = 0 0  0 1 1 a b 11 x y  a2  b2 Eqn. of line is a  b  1...........(1) P = distance of origin (0, 0) from (1) = 11 a2  b2 P2  1 1 1  1  P2  a2 b2 1  1 a2 b2 Midpoint D of BC = (a/2, 0); Midpoint of E of CA = (a/2, b/2) AD and BE are mutaully r , Slope of AD x Slope of BE = -1 0b x b 0   1; 2b2  a2 a  2b 2 0 a  0 a 2 2 9. 1 Midpoint of (5, -2) and (2, 2) = (7/2, 0); Any line r to 3x + 2y = 8 is 2x - 3y + k = 0 ..........(1) (1) passes through (7/2, 0); 2 x 7  2 x 0  k  0  k   7 ;Reqd. line is 2x - 3y - 7 = 0 2 10. 2 given line is x + y = 1 its slope = –1 ––––(1)  slope of any line to perpendicular it = 1 eqn of any line through (2,4) with slope 1 is y – 4 = 1(x–2)  x–y = –2  (2) 11. 2 12. 1 eqn (1) + eqn (2)  x = –1/2 eqn (1) – eqn (2)  y = 3/2 13. 2 If the required point is (0,y) then 0 12   y  22  0  32   y  44   4y  8y  20  0 Area = 0 ½[3(-3-a) = 2(a - 5) + 4(5 + 3)] = 0 -9 - 3a = 2a - 10 = 20 + 12 = 0; -a + 13 = 0 a = 13 Let 2x – 3y + k = 0 be the required line  y   2  x  k ;  y  int ercept is k ;  k  12  3  3 3

23V/M 2 J - MATHEMATICS 14. 1 All sides are equal and angles are 90° by puthagorous theorem 15. 3 p = a sin 600 or 1  a3 a  23 2 2 16. 4 17. 1 ; 18. 1 Take any two triplets satisfying a +b + c = 0 we can take a =2, b =-1, c = -1, then given equation 19. 2 becomes 4x - 3y -4=0 .........(1) and a =1, b =1, c = -2 equation becomes 2x + 3y -8 = 0 ..........(2). 20. 3  Solving equation 1 and 2 we get fixed point 2, 4 3 . 21. 3 The two given lines are parallel.Apoint on the line 3x + 4y = 9 is (3,0). The distance between the lines = distance of (3,0) from the line 6x + 8y – 15 = 0  6.3 8.0 15  3 /10 62  82 22. 45 23. 90 Here m1m2 = 2 (–1/2) = –1 24. 5 25. 2 Equation of the line joining the given point is y  4 6  4  x  3 , ie, y= x + 1. For the point where 5  3 coordinate is –1 ie y = –1, x = –1 –1 = –2