Problems on Locus and Introduction to Straight Lines

MATHEMATICS NOTE STRAIGHT LINES PROBLEMS ON LOCUS AND INTRODUCTION TO STRAIGHT LINES What you already know • Area of a triangle • Locus • Distance, section, midpoint formula • Different centres of a triangle What you will learn • Condition for lines to be parallel   and perpendicular • Angle of inclination • Slope of a line • Intercepts of a line •   Collinear lines •   Standard equation of a line • Angle between two lines Concept Check 1 Find the equation of the curve. (i) x = t2 + t + 1 and y = t2 - t + 1 ( ) ( )ett + e−−tt =and y (ii) x ett − e−−tt 22 (iii) x =t + 1 and y =t − 1 a tb t (iv) ax + by = 4t and axt - byt = 4 Find the locus of point P, whose ratio of distance from A(1, 2) and B(5, 6) is: (i) 1 (ii) 2 Solution (i) (h − 5)2 + (k − 6)2 According to the question, PA = 1 PB Let the locus of point P be (h, k ). PA = (h − 1)2 + (k − 2)2 and PB = (h − 1)2 + (k − 2)2 ⇒© 2020, BYJU'S. All rights reserved =1 22

02According to the question, PA = 1 (h − 5)2 + (k − 6)2 PB Let the locus of point P be (h, k ). PA = (h − 1)2 + (k − 2)2 and PB = (h − 1)2 + (k − 2)2 ⇒ =1 (h − 5)2 + (k − 6)2 ⇒ (h − 1)2 + (k − 2)2 = (h − 5)2 + (k − 6)2 Squaring and solving the above equation, and replacing (h, k ) with (x, y ) , we get, x + y =7, which is the locus of the point P. (ii) (h − 5)2 + (k − 6)2 We have, PA = 2 PB Let the locus of point P be (h, k ). PA = (h − 1)2 + (k − 2)2 and PB = (h − 1)2 + (k − 2)2 ⇒ =2 (h − 5)2 + (k − 6)2 ⇒ (h − 1)2 + (k − 2)2 = 2 (h − 5)2 + (k − 6)2 Squaring and solving the above equation, and replacing (h, k ) with (x, y ) , we get, 3x2 + 3y2 − 38x − 44y + 239 =0, which is a circle. Note 1. If a point P is such that for two given points, A and B,Wtheehraavtieo, PA ==12, then the locus of P is 2.  the perpendicular bisector of the line joining AB. PB k > 0), If a point P is such that for two given points, A and B,LWtehetethhraaevtlieoo,cPuAs of point P be (h, k ). ==k2, (where k ≠ 1 and then the locus of P is a circle. PA = (h P−B1)2 + (k − 2)2 and PB = (h − 5)2 + (k Let the locus of point P be (h, k ). (h −(h1−)21)+2 (k −(k2−)22)2=a2nd PB = (h − 5)2 + P⇒A = (k + (h − 5)2 + (k − 6)2 Concept Check 2 (h − 1)2 + (k − 2)2 If A ≡ (1, 2) and B ≡ (cos θ, sin θ), then find the locus of⇒⇒Stqhueap(r(hiohn−ig−n1t5a)t)n2h2da++tsoin((lkvtkein−−rgn26a)t)2lh2l=ye =22 (h − 5)2 + (k − 6)2 daibvoidveeseAqBuaintiothne, and replac ratio of 2 : 3. 3⇒x2 +(h3y−21−)238+x (k− −424)y2 =+ 2239(h=− 50),2w+hic(hk i−s6a)2circle Squaring and solving the above equation, and replac Concept Check 3 3x2 + 3y2 − 38x − 44y + 239 =0, which is a circle Find the locus of a point whose sum of distances from two perpendicular lines is 2 units. © 2020, BYJU'S. All rights reserved

03 Concept Check 4 Find the locus of a point that is equidistant from the x-axis and point (2, 3). Concept Check 5 Find the locus of the point S such that SQ2 + SR2 = 2SP2, where P ≡ (1, 0), Q ≡ (-1, 0), and R ≡ (2, 0). Master Problem A rod of length L slides along the coordinate axes such that its end points always lie on the x-axis and y-axis. Find the following: (a) Locus of the midpoint of the rod at all its positions (b) Locus of the circumcentre of the triangle (c) Locus of the centroid of the triangle made by the rod with the coordinate axes (d) Locus of a point that divides the rod’s length in the ratio of 1 : 2 that is measured from     the x-axis, where L = 4 Solution Y X B (b, 0) Let the end points of the rod be A and B, and A (a, 0) their coordinates be A (a, 0) and B (0, b). O (0, 0) Therefore, OA = a, OB = b, and the length of rod, AB = L As △OAB is a right-angled triangle, Using the Pythagoras theorem, we have, OA2 + OB2 = AB2 a2 + b2 = L2 (a) Let P (h, k) be the locus of midpoint of Y A (a, 0) X  rod AB.  By the midpoint formula, B (0, b) L h =(a + 0) and k (0 + b) b 22 P (h, k) ⇒ h =a and k = b Oa 22 =⇒ a 2h a=nd b 2k © 2020, BYJU'S. All rights reserved

04 Now, a2 + b2 = L2 Y A (a, 0) X ⇒ (2h)2 + (2k)2 = L2 ⇒ 4(h)2 + 4(k)2 = L2 B (0, b) Replace (h, k) by (x, y). Then, L 4x2 + 4y2 = L2 The locus of the midpoint of the rod is a circle. b P (h, k) (b) We have to find the locus of the circumcentre Oa of ΔOAB. ΔOAB is a right-angled triangle. In a right-angled triangle, the midpoint of the hypotenuse is the circumcentre of the triangle. We already know from the previous solution that the locus of the midpoint of the hypotenuse is 4x2 + 4y2 = L2 As the midpoint is the circumcentre itself, the locus is the same circle, 4x2 + 4y2 = L2 (c) Y Let G(h, k) be the locus of centroid of ΔOAB. We know that the vertices of ΔOAB are O(0, 0), B (0, b) A (a, 0), and B (0, b). By the centroid formula, L b (h, k) ≡  (0 + a + 0) , (0+0+ b)   3  G (h, k)  3  Oa ⇒ h =a and k = b 33 =⇒ a 3h a=nd b 3k Now, we have a2 + b2 = L2 A (a, 0) X ⇒ (3h)2 + (3k)2 = L2 ⇒ 9(h)2 + 9(k)2 = L2 Replace (h, k) by (x, y). Then, 9x2 + 9y2 = L2 ⇒ x2 + y2 = L2 . 9 The locus of the centroid of the triangle is a circle. © 2020, BYJU'S. All rights reserved

05 (d) Let P(h, k) be the locus point which lies on the rod and divides the rod’s length in the ratio of 1 : 2 that is measured from the x-axis, where L = 4 By the section formula of internal division, Y (h, k) ≡  (0 + 2a) , (b+0)     3 3  (0, b) ⇒ h =2a and k = b 2 33 P (h, k) 1 =⇒ a 3h =and b 3k. O (a, 0) 2 Now, a2 + b2 =L2 X ⇒  3h 2 + (3k )2 =L2  2  ⇒ 9 (h)2 + 9(k )2 = L2 4 Replace (h, k ) by (x, y ) and L = 4. Then, 9 x2 + 9y2 = 42 4 Concept Check 6 Find the locus of a point that moves such that its distance from point (0, 0) is twice the distance from the y-axis. a) 3x2 - y = 2 b) 3x2 - y2 = 0 c) x2 + 3y2 = 2 d) x2 + 2y2 = 2 Angle of Inclination Y It is an angle theta (θ) that a line makes O 0≤θ<π with the positive direction of x-axis which θ is measured in the anticlockwise sense. Also, θ ∈ [0, π) X © 2020, BYJU'S. All rights reserved

06 Note Y l1 l2 1) Parallel lines have the same angle θ1 = θ2  of inclination. θ1 θ2 X O 2) The angle of inclination of a line Y θ = 0° parallel or coincident with the l1 X x-axis is 0. O l2 3) The angle of inclination of a line parallel Y θ = 90° or coincident with the y-axis is π . l2 l1 2 O X © 2020, BYJU'S. All rights reserved

07 Slope of a Line Y If the angle of inclination of a given line l, l θ is θ, then the slope, m, of that line is given O by tan θ. X Example Y l When we are given a straight line making an angle of 30° with the positive direction 30° X of x-axis then, the slope of the line is, O =tan θ ta=n 300 1 Y X 3 l Therefore, m = 1 120° 3 O If the angle of inclination of the line θ is 60°, then the slope is tan 60° = 3 Now, if the angle of inclination is 120°, then the slope is tan 120° = -√3 If we are given that slope m = -√3, what would be the value of θ? tan θ = -√3 ⇒ tan θ = -tan 60° ⇒ tan θ = tan (180° - 60°) ⇒ θ = 180° - 60° = 120° © 2020, BYJU'S. All rights reserved

08 Note Y 1) Two parallel lines have same slope, i.e; O l1 l2  ml1 = ml2 and vice versa. θ1 ml1 = ml2 θ2 X 2) If 0 <θ < π , then m = tan θ > 0. The slope Y m = tan θ > 0 2 X is positive. θ O 3) If θ= π , ⇒ m = tan π , which is Y 2 2 θ = 90° X undefined. O 4) If π < θ < π, then m = tan θ < 0. The slope Y m = tan θ < 0 2 X is negative. θ O © 2020, BYJU'S. All rights reserved

09 Concept Check 7 Find the angle of inclination of the line whose slope is 1 . −1 33 (y2 − y1 ) (x2 − x1 ) Concept Check 8 Find the angle of inclination of the line whose slope1is −1 . 33 (y2 − y1 ) (x2 − x1 ) Calculation of Slope 1 −1 33 The slope (m) of the line joining points P (x1, y1) and Q (x2, y2) is given by (y2 − y1 ) . (x2 − x1 ) Proof Drop perpendiculars from points P and Q Y , y ) onto the x-axis and also draw a horizontal y2 line (parallel to the x-axis) that is passing 2 2 through point P and intersects the y1 perpendicular dropped from point Q at N. (x Q A right-angled triangle, △PNQ, is formed. Let the angle of inclination of the line be θ. (x , y ) y2 - y1 Therefore, the slope will be tan θ. N P 1 1 θ From ∆PNQ, tanθ = QN x2 - x1 PN θ x1 x2 X ⇒=tan θ ((=yx22 −− yx11 )) m O Therefore, the slope of the line joining points P and Q is (y2 − y1 ) . (x2 − x1 ) Find the slope of a line joining points (2, 1) and (0, -3). (a) 1 (b) -2 (c) 1 (d) 2 Solution Let the points given be (x1, y1) ≡ (2, 1) and (x2, y2) ≡ (0, -3) © 2020, BYJU'S. All rights reserved

10 Slo=pe, m ((yx22=−− yx11 )) =−3 − 1 2 0−2 OOppttiioonn ((dd)) iiss tthheeccoorrrreeccttaannsswweerr.. Condition for Collinearity lY If points A, B, and C are collinear, then the slope A of line AB is the same as the slope of line BC and mAB AC, which means mAB = mBC = mAC. B mBC O X C Find x if the points A (2, 3), B (1, 1), and C (x, 3x) are collinear. (a) 1 (b) -1 (c) 1 (d) 2 Solution If points A, B, and C are collinear, then the slope of line AB is the same as the slope of line BC, which means mAB = mBC =mAB =1 − 3 2 and=mBC 3x − 1 1−2 x −1 ⇒ 2 =3x − 1 x −1 ⇒ x =− 1 OOppttiioonn (b) is thhee ccorrrreeccttaannsswweerr.. © 2020, BYJU'S. All rights reserved

11 Angle between Two Lines Let θ be the acute angle between lines l1 and θl22h, raevsinpgecatnivgelely.of inclinations as θ1 and Y l1 Slope of l1 = m1 = tan θ1 π-θ l2 Slope of l2 = m2 = tan θ2 θ X θ1 By the property of a triangle, the sum of two θ2 interior angles is equal to the exterior angle O opposite to them. Htweorein, tθe1riisorthaengelxetserfioorrtahnegtlreia, nθg2 laenhdigθhalirgehttheed in the figure. T⇒heθr=efθo1r-eθ, θ2 1 = θ + θ2 Applying tangent function on both the sides, ⇒ tan=θ tan (θ1 − θ2 ) ⇒ tan θ = tan θ1 − tan θ2 1 + tan θ1 tan θ2 ⇒ tan θ = m1 − m2 1 + m1m2 The acute angle between the lines is given by, tan θ = m1 − m2 1 + m1m2 Note • l1 ‖ l2 ⇒ m1 = m2 • m1 . m2 = -1 ⇒ l1 ⟂ l2, but the converse is not true, i.e., if the lines are perpendicular, then it is not necessary that the product of their slopes is -1. Example: Say, l1 : x = 4 and l2 : y = 2 lli1niessthweitlhintehepiarrsallolepletso my-1a=xiusnadnedfiln2eisd the parallel to x-axis and hence these are perpendicular and m2 = 0, but the product is not -1. • If m1 × m2 = 1 ⇒ Lines are at a complementary angle with each other (θ1 + θ2 = 90°) Concept Check 9 Find the locus of point P, whose ratio of distance from A(1, 2) and B(5, 6) is 1. © 2020, BYJU'S. All rights reserved

12 Intercepts of a Line The intercept of a line is the point at which it crosses either the x-axis or the y-axis. An intercept can be negative, positive, or zero. Given are the intercepts of some lines. YY B (0, 3) B (0, 3) x+y=3 -x + y = 3 O A (3, 0) X A (-3, 0) O X x-intercept : OA = 3 x-intercept : OA = -3 y-intercept : OB = 3 y-intercept : OB = 3 YY A (-3, 0) O X O X x + y = -3 B (0, -3) B (0, -3) A (3, 0) x-y=3 OA = -3 OB = -3 x-intercept : OA = 3 x-intercept y-intercept y-intercept : OB = -3 Note Both the x-intercept and the y-intercept will be zero if the line passes through the origin (0, 0). © 2020, BYJU'S. All rights reserved

13 Standard Equation of a Line We know that the standard equation of the line is y = mx + c, where m and c are real constants. For a line parallel to the x-axis, the slope will be zero, i.e., m = 0 ⇒ y = c or y = k (where k is any real number) Similarly, for a line parallel to the y-axis, slope m is undefined. ⇒ x = k (where k is any real number) Summary Sheet Key Takeaways • If a point P is such that for two given points, A and B, thWe erahtiaovoef, PA ==12, then the locus of P is the perpendicular bisector of the line joining AB. PB • If a point P is such that for two given points, A and B, thLWeeertaththiaoevoleof,cPuAs o=f=kp2o(winht ePrbeek (≠h1, kan).d k > 0), LPeAt t= he l(ohcuP−sB1)o2f p+oi(nkt −P2b)e2 a(hn,dkP)B. = (h − 5)2 + then the locus of P is a circle. Key Definitions ⇒PA = (h −(h1−)21)+2 (k −(k2−)22)2=a2nd PB = (h − 5)2 + of inclination: It is an angle theta (θ) that a lθin∈e[m0,aπk⇒⇒)e. s measured in the anticlockwise sense. Also, + w(i(((thhhhh−−−−t1h551)e)))2222p+o+++s(i(((tkkkkiv−−−−e2662d))))2i222r=ec=ti2o2n(ohf−t5h)e2 + (k − 6)2 Angle 3⇒Sθxq, 2tuh+ae(rnhi3ny−gth21ae−)n2s3dlo8+sxpoe(lkv−,imn−4g2,4o)ty2hf=et+haa2b2t3oli9vn(hee=−eis5q0)u,2awt+ihoinc(h,kai−sn6da)2rceirp x-axis Slope of a line: If the angle of inclination of a given line is given⇒byttaan=n θθ. tan θ2 (θ1 − ) ⇒ tan θ = tan θ1 − tan θ2 Squaring and solving the above equation, and rep Key Form1ul+aetan θ1 tan θ2 3x2 + 3y2 − 38x − 44y + 239 =0, which is a cir • Th⇒e stlaonpeθ(m=) o1mf t+1h−emlm1inm2e2 joining points P(x1, y1) and Q(x2, y2)Silso=pgeiv,emn by ((yx22=−− yx11 )) . =−3 − 1 2 • ThTeheacauctueteanagnlgel,eθbbeetwtweeeennththeelilnineessisisggiviveennbbyy, ttaann θθ = 1mO+1p−mtiom1nm2(2d). is 0−2 the correct answer. Key Results • Parallel lines have the same angle of inclination. • The angle of inclination of a line parallel or coincident with the x-axis is 0. • The angle of inclination of a line parallel or coincident with the y-axis is π . 2 If θ is the angle of inclination of a line, • 0<θ< π , then m = tan θ > 0. The slope is positive. 2 θ2π= <π2 θ⇒< π • If m = tan 2 , which is undefined. is negative. • If π, then m = tan θ < 0. The slope © 2020, BYJU'S. All rights reserved

14 If θ is the angle between lines L1 and L2 whose slopes are m1 and m2 respectively, • L1 ‖ L2 ⇒ θ = 0 ⇒ m1= m2 • ims 1n.omt 2n=ec-1es⇒saLry1 ⟂thLa2t , but the converse is not true, i.e., if the lines are perpendicular, then it the product of their slopes is -1. • If m1 × m2 = 1 ⇒ Sum of the angle of inclination of the lines is 90°. Condition for collinearity • If points A, B, and C are collinear, then the slope of line AB is the same as the slope of line BC and AC, which means mAB = mBC = mAC Intercept of a line • The intercept of a line is the point at which it crosses either the x-axis or the y-axis. • An intercept can be negative, positive, or zero. • If the line passes through the origin, then both the x-intercept and the y-intercept will be zero. Standard equation of a line parallel to the coordinate axes • A line parallel to the x-axis will be in the form of y = k (where k is any real number) • A line parallel to the y-axis will be in the form of x = k (where k is any real number) Mind Map Angle of inclination Straight Slope line Collinear lines Angle between two lines Intercepts of a line Lines parallel to coordinate axes Self-Assessment Find the equation of a line that is parallel to X-axis and passing through the point (5, -6). © 2020, BYJU'S. All rights reserved

15 Answers Concept Check 1 (i) Given, x = t2 + t + 1 ; y = t2 - t + 1 (x − y) ⇒ =t (x −2 y) S==⇒⇒⇒ubxxstit2utinxxg−−22=thyyte22va++luexx o−−22fyyt in one of the equations, + 1 + 1 ⇒ 4x = x2 + y2 - 2xy + 2x - 2y + 4 ⇒ x2 + y2 - 2xy - 2x - 2y + 4 = 0 (ii) ( )et + e−t ( )et − e−t =Given, x =; y 2 2 On squaring x and y, we get, ( ) ( )e2t + e−2t + 2 x2 =; y2 e2t + e−2t − 2 44 ( )⇒ 4x2 − 2 = 4y2 + 2 Eliminating e2t + e−2t ( )⇒ 4 x2 − y2 =4 ( )⇒ x2 − y2 =1 (iii) Given, x =t + 1 , y =t − 1 a tb t Squaring x and y , we get, ab x2 =t2 + 1 + 2 and y2 =t2 + 1 −2 a2 t2 b2 t2 ⇒ x2 − 2= y2 + 2  Eliminating t2 + 1  a2 b2  t2  ⇒ x2 − y2 =4 a2 b2 (iv) Given, axt - byt = 4 ⇒ ax - by = 4 and ax + by = 4t t Now, multiplying both the equations to eliminate t, we get, (ax - by)(ax + by) = 16 ⇒ a2 x2 - b2 y2 = 16 © 2020, BYJU'S. All rights reserved

16 Concept Check 2 Given, A(1, 2) and B(cos θ, sin θ) Let P(h, k) be the locus of the point that divides line AB in the ratio of 2:3. By using the section formula, h =(2cosθ + 3) , k (2sinθ + 6) 55 =cos θ (5h=− 3) , sinθ (5k − 6) 22 ⇒  (5h − 3) 2  (5k − 6) 2  +  =1 2  2  ⇒  (5x − 3) 2  (5y − 6) 2 =1 (Replacing (h, k) with (x, y ))  +    2   2  Concept Check 3 Y Without loss of generality we can assume P (h, k) |h| perpendicular lines to be the x and y axes. Let P(h, k) be the locus of the point. |k| Distance of P from the x-axis = |k| And from the y-axis = |h| O X Therefore, we get |h| + |k| = 2 Hence, the locus is |x| + |y| = 2 Concept Check 4 Let the locus of the point be (h, k) Distance from the x-axis = |k| Distance from (2, 3) is (h - 2)2 + (k - 3)2 (h - 2)2 + (k - 3)2 = k2 (Equating the distances and squaring on both sides) Hence, x2 - 4x - 6y + 13 = 0 is the required locus. (Solving and replacing (h, k) with (x, y)) Concept Check 5 Let locus of S be (h, k). Then, SQ2 + SR2 = 2SP2 (Calculating distances, substituting in the above equation) ⇒ (h + 1)2 + (k)2 + (h - 2)2 + (k)2 = 2((h - 1)2 + (k)2) ⇒ 2h - 4h + 5 = -4h + 2 ⇒ h = -3 2 Required locus is obtained by replacing h with x, which is x = -3 2 © 2020, BYJU'S. All rights reserved

17 Concept Check 6 Y P (h, k) M X Let P(h, k) be the moving point whose locus is required. O B Given, |OP| = 2|PM| ⇒ h2 + k2 = 2|h| P ⇒ h2 + k2 = 4h2 ⇒ 3h2 - k2 = 0 Replacing (h, k) with (x, y), we get the required locus as 3x2 - y2 = 0 So, option (b) is the correct answer. Concept Check 7 Since, m= tanθ= 1 3 ⇒ tanθ =tan π 6 ⇒ θ =π 6 Concept Check 8 Given, slope of the line, m= tanθ= −1 3 ⇒ tanθ = − tan π 6 ⇒ ta=nθ tan  π − π  6  ⇒ tanθ =tan 5π 6 Hence, the angle of inclination of the line is 150°. Concept Check 9 PA = 1 PB HoUMM⇒PPfseABiilddininnS=ppcgelooe1otii,sphnnPeeettgooloismeffefsAAecAnoBBtBitonAn≡≡=thBfoe.((r((65pm11−e−22++ur12lp55a))e)), ,,n=d((22i1c++22ul66a))r bis≡ec(t3o, r4)   ≡ (3, 4) A   ( )⇒ Slope of AB =6 − 2 =1 ( )5 − 1© 2020, BYJU'S. All rights reserved

18 PA = 1 PB Midpoint poef rApBen≡dicu(l1a2+r b5i)s,ec(2to+2r 6)  (3, 4) to AB itself. Also, the ≡ of AB is perpendicular ⇒ Slope of AB =((65 −− 2) =1 1) ⇒ The slope of the perpendicular bisector of AB is -1. Therefore, the equation of the perpendicular bisector of AB passing through (3, 4) and the slope of -1 is y - 4 = -1(x - 3) ⇒ x + y = 7, which is the locus of P. Self-Assessment Let the line that is parallel to X-axis be L. We know that equation of line parallel to X-axis is given by y=k Given that line passes through (5, -6), ⇒ y = -6, is the required equation. © 2020, BYJU'S. All rights reserved