Pair of Straight Lines PDF

3.1 Equation of Pair of Straight lines. Let the equation of two lines be a x  b y  c   0.....(i) and a x  b y  c   0 60 60 Pair of Straight Lines.....(ii) E3 Hence (a x  b y  c )(a x  b y  c )  0 is called the joint equation of lines (i) and (ii) and conversely, if joint...

3.1 Equation of Pair of Straight lines. Let the equation of two lines be a x  b y  c   0.....(i) and a x  b y  c   0 60 60 Pair of Straight Lines.....(ii) E3 Hence (a x  b y  c )(a x  b y  c )  0 is called the joint equation of lines (i) and (ii) and conversely, if joint equation of two lines be (a x  b y  c ) (a x  b y  c )  0 then their separate equation will be a x  b y  c   0 and ax  by  c  0. (1) Equation of a pair of straight lines passing through origin : The equation ax 2  2hxy  by 2  0 ID represents a pair of straight line passing through the origin where a, h, b are constants. Let the lines represented by ax 2  2hxy  by 2  0 be y  m 1 x  0 and y  m 2 x  0  h  h 2  ab  h  h 2  ab 2h a and m 2 = then, m 1  m 2   and m 1 m 2  b b b b U where, m 1  Then, two straight lines represented by ax 2  2hxy  by 2  0 are ax  hy  y h 2  ab = 0 and Note : D YG ax  hy  y h 2  ab  0.  The lines are real and distinct if h 2  ab  0  The lines are real and coincident if h 2  ab  0  The lines are imaginary if h 2  ab  0  If the pair of straight lines ax 2  2hxy  by 2  0 and a' x 2  2h' xy  b' y 2  0 should U have one line common, then (ab'a' b) 2  4(ah'a' h)(hb'h' b).  The equation of the pair of straight lines passing through origin and perpendicular to the pair of straight lines represented by ax 2  2hxy  by 2  0 is ST given by bx 2  2hxy  ay 2  0  If the slope of one of the lines represented by the equation ax 2  2hxy  by 2  0 be the square of the other, then a 2 b  ab 2  6 abh  8 h 3  0.  If the slope of one of the lines represented by the equation ax 2  2hxy  by 2  0 be  times that of the other, then 4 h 2  ab(1  )2. (2) General equation of a pair of straight lines : An equation of the form, ax 2  2hxy  by 2  2 gx  2 fy  c  0 where a, b, c, f, g, h are constants, is said to be a general equation of second degree in x and y. Pair of Straight Lines 61 The necessary and sufficient condition for ax 2  2hxy  by 2  2 gx  2 fy  c  0 to represent a a h pair of straight lines is that abc  2 fgh  af  bg  ch  0 or h b g f 2 2 g f 0 c 2 60 (3) Separate equations from joint equation: The general equation of second degree be ax  2hxy  by 2  2 gx  2 fy  c  0. To find the lines represented by this equation we proceed as 2 follows : Step I : Factorize the homogeneous part ax 2  2hxy  by 2 into two linear factors. Let the E3 linear factors be a' x  b ' y and a" x  b " y. Step II: Add constants c ' and c" in the factors obtained in step I to obtain a' x  b ' y  c' and a" x  b " y  c". Let the lines be a' x  b ' y  c'  0 and a" x  b " y  c"  0. If the sum of the slopes of the lines given by x 2  2cxy  7 y 2  0 is four times their product. Then c has U Example: 1 ID Step III : Obtain the joint equation of the lines in step II and compare the coefficients of x, y and constant terms to obtain equations in c' and c". Step IV : Solve the equations in c' and c" to obtain the values of c' and c". Step V : Substitute the values of c' and c" in lines in step II to obtain the required lines. the value Solution: (c) (b) – 1 We know that, m 1  m 2  (b) bm 2  2hm  a  0 (c) am 2  2hm  b  0 U ax 2  2hx (mx )  b(mx )2  0  a  2hm  bm 2  0 If the equation 12 x 2  10 xy  2y 2  11 x  5 y  K  0 represent two straight lines, then the value of K is [MP PET 20 ST (a) 1 Solution: (b) (d) bm 2  2hm  a  0 Substituting the value of y in the equation ax 2  2hxy  by 2  0  Example: 3 2c  1   4  c  2 7 7 If one of the lines represented by the equation ax 2  2hxy  by 2  0 be y  mx , then (a) bm 2  2hm  a  0 Solution: (a) (d) 1 2h a and m 1 m 2 . b b Given, m1  m2  4 m1m2  Example: 2 [AIEEE 2004] (c) 2 D YG (a) – 2 Condition (b) 2 for pair (c) 0 of lines, (d) 3 abc  2 fgh  af 2  bg 2  ch 2  0 , Here a  12, h  5, b  2, g  11 / 2, f  5 / 2, c  K Then, 12  2  K  2  2 2  5 11 5  11    12     2     K(5) 2  0. On solving, we get K= 2. 2 2  2   2  3.2 Angle between the Pair of Lines. (1) The angle  between the pair of lines represented by ax 2  2hxy  by 2  0 is given by tan   2 h 2  ab ab (i) The lines are coincident if the angle between them is zero. 62 Pair of Straight Lines Lines are coincident i.e.,   0  tan   0   2 h 2  ab  0  h 2  ab  0  h 2  ab ab Hence, the lines represented by ax 2  2hxy  by 2  0 are coincident, iff h 2  ab (ii) The lines are perpendicular if the angle between them is  / 2.   cot   cot 2  2 ab  cot   0  2 h  ab 2  0  a  b  0  coeff. of x 2  coeff. of 60   y 0 2 of y 2  0. E3 Thus, the lines represented by ax 2  2hxy  by 2  0 are perpendicular iff a  b  0 i.e., coeff. of x 2  coeff. (2) The angle between the lines represented by ax 2  2hxy  by 2  2 gx  2 fy  c  0 is given by 2 h 2  ab ab    tan 1 2 h 2  ab ab ID tan   (i) The lines are parallel if the angle between them is zero. Thus, the lines are parallel iff 2 h 2  ab  0  h 2  ab. ab U   0  tan   0  D YG Hence, the lines represented by ax 2  2hxy  by 2  2 gx  2 fy  c  0 are parallel iff h 2  ab and af 2  bg 2 or a h g  . h b f (ii) The lines are perpendicular if the angle between them is  / 2. ab Thus, the lines are perpendicular i.e.,    / 2  cot   0  0 2 h 2  ab  a  b  0  coeff. of x 2  coeff. of y 2  0 ax 2  2hxy  by 2  2 gx  2 fy  c  0 are perpendicular iff U Hence, the lines represented by ab  0 i.e., coeff. of x 2  coeff. of y 2  0. ST (iii) The lines are coincident, if g 2  ac. Example: 4 The angle between the lines x 2  xy  6 y 2  7 x  31 y  18  0 is (a) 45 o (b) 60 Angle between the lines is   tan 1 45 o Example: 5 If the angle between the x  3 xy  y  3 x  5 y  2  0 is tan (a) 2 (c) 90 o 2 (b) 0 [Karnataka CET 2003] (d) 30 2 Solution: (b) 2 o o  1 2 6 2    1  (6) 2 h 2  ab  4  2  = tan 1  tan 1 | 1 |  tan 1 (1)  , 1  tan ab 4 1  (  6 ) 1  (6 ) 1 pair of straight lines represented by the 1   , where  is a non- negative real number, then  is 3 1  (c) 3 (d) 1 equation Pair of Straight Lines 63 1 1 Given that   tan 1    tan   3 3 Solution: (a) 2  (  1) 2  9(9  4 )  2  38   80  0 60 3 2    1  2  2 h 2  ab Now, since tan    =  1 3 ab  2  40   2  80  0  (  40 )  2(  40 )  0  (  2)(  40 )  0    2 or – 40, but  is a nonnegative real number. Hence   2. The angle between the pair of straight lines represented by 2 x 2  7 xy  3 y 2  0 is Example: 6 (a) 60 o E3 [Kurukshetra CEE 2002] (c) tan 1 (7 / 6) (b) 45 o (d) 30 o 2 Angle between the lines is ,   tan 1 2 h 2  ab  tan 1 ab 2 5    tan 1 .   tan 1 (1)  5 2 ID Solution: (b)  7 2     (2)(3)  2 23   45 o 3.3 Bisectors of the Angles between the Lines.  x 2  y 2 xy  ab h.....(i) D YG equation ax 2  2hxy  by 2  0 is U (1) The joint equation of the bisectors of the angles between the lines represented by the hx 2  (a  b)xy  hy 2  0 Here, coefficient of x 2  coefficient of y 2  0. Hence, the bisectors of the angles between the lines are perpendicular to each other. The bisector lines will pass through origin also. Note : If a  b , the bisectors are x 2  y 2  0 i.e., x  y  0, x  y  0  If h  0 , the bisectors are xy  0 i.e., x  0, y  0. U  If bisectors of the angles between lines represented by ax 2  2hxy  by 2  0 and h' a'b '.  h ab ST a' x 2  2h' xy  b ' y 2  0 are same, then  If the equation ax 2  2hxy  by 2  0 has one line as the bisector of the angle between the coordinate axes, then 4 h 2  (a  b)2. (2) The equation of the bisectors of the angles between the lines represented by (x   )   (y   ) 2 (x   )(y   )  , where ,  is the ab h point of intersection of the lines represented by the given equation. ax 2  2hxy  by 2 + 2 gx  2 fy  c  0 are given by Example: 7 The equation of the bisectors of the angles between the lines represented by x 2  2 xy cot   y 2  0 is (a) x 2  y 2  0 (b) x 2  y 2  xy x y xy x y xy   or 0 cot  ab h 2 Solution: (a) Equation of bisectors is given by (c) (x 2  y 2 ) cot   2 xy 2 2 2 (d) None of these  x2  y2  0 64 Pair of Straight Lines Example: 8 If the bisectors of the lines x 2  2 pxy  y 2  0 be x 2  2qxy  y 2  0, then [MP PET 1993; DCE 1999; Rajasthan PET 2003; AIEEE 2003] Solution: (a) (b) pq  1  0 (c) p  q  0 Bisectors of the angle between the lines x 2  2 pxy  y 2  0 is But it is represented by x 2  2qxy  y 2  0. Therefore (d) p  q  0 x2  y2 1  (1)   px 2  2 xy  py 2  0 xy p p 2   pq  1  pq  1  0 1  2q 60 (a) pq  1  0 For point of intersection (Keeping y as constant) (Keeping x as constant)   0  0 and y x ID and   2ax  2hy  2 g  0 x   2 hx  2by  2 f  0 y E3 3.4 Point of Intersection of Lines represented by ax2+2hxy+by2+2gx+2fy+c = 0. Let   ax 2  2hxy  by 2  2 gx  2 fy  c  0 We obtain, ax  hy  g  0 and hx  by  f  0 x y 1   fh  bg gh  af ab  h 2 U On solving these equations, we get  bg  fh af  gh  , 2  2  h  ab h  ab  i.e. (x , y )   D YG a h g Also, since   h b f , from first two rows g f c a h g  ax  hy  g  0 and h b f  hx  by  f  0 and then solve, we get the point of intersection. Note The point of intersection of the lines represented by the equation 2 x 2  3 y 2  7 xy  8 x  14 y  8  0 is U Example: 9 :  The point of intersection of lines represented by ax 2  2hxy  by 2  0 is (0, 0). (a) (0 , 2) (c) (2,0) (d) (2, 1) Let   2 x 2  3 y 2  7 xy  8 x  14 y  8  0 ST Solution: (c) (b) (1, 2)    4 x  7 y  8  0 and  6 y  7 x  14  0 x y On solving these equations, we get x  2, y  0 Trick : If the equation is ax 2  2hxy  by 2  2 gx  2 fy  c  0  hf  bg hg  af  , The points of intersection are given by . Hence point is (– 2, 0) 2 2  ab  h ab  h  Example: 10 If the pair of straight lines xy  x  y  1  0 and line ax  2 y  3  0 are concurrent, then a = (a) – 1 Solution: (d) (b) 0 (c) 3 Given that equation of pair of straight lines xy  x  y  1  0 (d) 1 Pair of Straight Lines 65  (x  1)(y  1)  0  x 1  0 or y  1  0 The intersection point of x  1  0, y  1  0 is (1,1)  Lines x  1  0, y  1  0 and ax  2 y  3  0 are concurrent.  The intersecting points of first two lines satisfy the third line. 60 Hence, a  2  3  0  a  1 3.5 Equation of the Lines joining the Origin to the Points of Intersection of a given Line and a given Curve. The equation of the lines which joins origin to the point of intersection of the line homogeneous with the help of 2  lx  my   lx  my  ax 2  2hxy  by 2  2(gx  fy)   c  0  n   n  and line lx  my  n  0 , ID We have ax 2  2hxy  by 2  2 gx  2 fy  c  0 E3 lx  my  n  0 and curve ax 2  2hxy  by 2  2 gx  2 fy  c  0 , can be obtained by making the curve lx  my  n  0 which is......(i).....(ii) U Suppose the line (ii) intersects the curve (i) at two points A and B. We wish to find the combined equation of the straight lines OA and OB. Clearly OA and OB pass through the origin, so their joint equation is a homogeneous D YG equation of second degree in x and y. From equation (ii), lx  my  n......(iii)  Y A lx+my+n=0 lx  my 1 n X' B O X Now, consider the equation 2  lx  my   lx  my   lx  my  ax  2hxy  by  2 gx   0.....(i   2 fy   c  n   n   n  2 U v) Y 2 ST Clearly, this equation is a homogeneous equation of second degree. So, it represents a pair of straight lines passing through the origin. Moreover, it is satisfied by the points A and B. Hence (iv) represents a pair of straight lines OA and OB through the origin O and the points A and B which are points of intersection of (i) and (ii). Example: 11 The lines joining the origin to the point of intersection of the circle x 2  y 2  3 and the line x  y  2 are (a) y  (3  2 2 )x  0 (b) x  (3  2 2 )y  0 (c) x  (3  2 2 )y  0 (d) y  (3  2 2 )x  0 Solution: (a,b,c,d) Make homogenous the equation of circle, we get x 2  6 xy  y 2  0  x 6 y  (36  4 )y 2 2  6y  4 2y  3y  2 2y 2 Hence, the equation are x  (3  2 2 )y and x  (3  2 2 )y Also after rationalizing these equations becomes y  (3  2 2 )x  0 and y  (3  2 2 )x  0. 66 Pair of Straight Lines Example: 12 The pair of straight lines joining the origin to the points of intersection of the line y  2 2 x  c and the circle x 2  y 2  2 are at right angles, if [MP PET 1996] (a) c 2  4  0 Pair of straight lines joining the origin to the points of intersection of the line y  2 2 x  c and the circle x 2  y 2  2 are  2  60 Solution: (c) (d) c 2  10  0 (c) c 2  9  0 (b) c 2  8  0  2 2x  y    0  x 2  y 2  2 8 x 2  y 2  4 2 xy  0  x 2  1  16   y 2  1  2   8 2 xy  0 x  y  (2)    c c2  c2  c2 c2     2 If these lines are perpendicular, 1  2c 2  18  c 2 0  c 2 16 2 1  2  0 c2 c E3  2 9  0. ID 3.6 Removal of First degree Terms. ax 2  2hxy  by 2  2 gx  2 fy  c  0......(i) Let point of intersection of lines represented by is ( ,  ). D YG be (Y   ) in (i). U  bg  fh af  gh  Here ( ,  )   2 , 2   h  ab h  ab  For removal of first degree terms, shift the origin to ( ,  ) i.e., replacing x by ( X   ) and y Alternative Method : Direct equation after removal of first degree terms is aX 2  2hXY  bY 2  (g  f  c)  0 Where   bg  fh af  gh and   2 2 h  ab h  ab 3.7 Removal of the Term xy from f (x, y) = ax2 + 2hxy +by2 without changing the Origin.  U Clearly, h  0. Rotating the axes through an angle  , we have, x  X cos   Y sin  and y  X sin   Y cos  f (x , y)  ax 2  2hxy  by 2 ST After rotation, new equation is F(X , Y )  (a cos 2   2h cos  sin   b sin 2  )X 2  2{(b  a)cos  sin   h(cos 2   sin 2  )XY  (a sin 2   2h cos  sin   b cos 2  )Y 2 ab Now coefficient of XY = 0. Then we get cot 2  2h Note :  Usually, we use the formula, tan 2  . However, if a  b , we use cot 2  Example: 13 2h ab for finding the angle of rotation, ab as in this case tan 2 is not defined. 2h The new equation of curve 12 x 2  7 xy  12 y 2  17 x  31 y  7  0 after removing the first degree terms (a) 12 X 2  7 XY  12Y 2  0 (b) 12 X 2  7 XY  12 Y 2  0 Pair of Straight Lines 67 (c) 12 X 2  7 XY  12Y 2  0 Solution: (c) (d) None of these Let   12 x  7 xy  12 y  17 x  31 y  7  0 2 2.....(i)    24 x  7 y  17  0 and  7 x  24 y  31  0 x y  Their point of intersection is (x , y )  (1,1) 60 Here   1,   1 Shift the origin to (1, –1) then replacing x  X  1 and y  Y  1 in (i), the required equation is 12(X  1)2  7(X  1)(Y  1)  12(Y  1)2  17 (X  1) 31(Y  1)  7  0 i.e., 12 X 2  7 XY  12Y 2  0 Alternative Method : Here   1 and   1 and g  17 / 2, f  31 / 2, c  7 17 31 1   1  7  0 2 2 g  f  c    Removed equation is aX 2  2hXY  bY 2  (g  f  c)  0 12 X 2  7 XY  12 Y 2  0  0  12 X 2  7 XY  12Y 2  0. i.e., Example: 14 E3  Mixed term xy is to be removed from the general equation ax 2  by 2  2hxy  2 gx  2 fy  c  0 , one should ab 2h (a) Solution: (d) (b) 2h ab ID rotate the axes through an angle  given by tan 2 = (c) ab 2h (d) 2h ab Let (x ' , y ' ) be the coordinates on new axes, then put x  x ' cos   y ' sin  , y  x ' sin   y ' cos  in the D YG U equation, then the coefficient of xy in the transformed equation is 0. 2h So, 2(b  a) sin. cos   2h cos 2  0  tan 2  ab 3.8 Distance between the Pair of parallel Straight lines s If ax 2  2hxy  by 2  2 gx  2 fy  c  0 represent a pair of parallel straight lines, then the distance between them is given by 2 Example: 15 Distance between the pair of lines represented by the equation x 2  6 xy  9 y 2  3 x  9 y  4  0 [Kerala (Engg.) 20 15 1 5 (b) (c) 2 2 10 The distance between the pair of straight lines given by (a) (d) U Solution: (c) g 2  ac f 2  bc or 2 b(a  b) a(a  b ) g 2  ac 3 ax  2hxy  by  2 gx  2 fy  c  0 is 2 , Here a  1, b  9, c  4 , g   2  a(a  b ) 2 2 ST 2 Example: 16 Solution: (a) 1 10 9  (4 ) 4  2 1(1  9 ) 25 4  10 5 2 Distance between the lines represented by the equation x 2  2 3 xy  3 y 2  3 x  3 3 y  4  0 is [Roorkee 1989] (a) 5/2 (b) 5/4 (c) 5 (d) 0 3 1 3 a h g 2   First check for parallel lines i.e.,    3 h b f 3 3 3 2 which is true, hence lines are parallel.  5/2 3.9 Some Important Results Distance between them is 2 (3 / 2)2  1(4 ) g 2  ac 2 a(a  b ) 1(1  3) 68 Pair of Straight Lines (1) The lines joining the origin to the points of ax  2hxy  by 2  2 gx  0 and a' x 2  2h' xy  b ' y 2  2 g' x  0 will be g(a'b ' )  g ' (a  b). 2 intersection of the curves mutually perpendicular, if (2) If the equation hxy  gx  fy  c  0 represents a pair of straight lines, then fg  ch. 60 (3) The pair of lines (a 2  3b 2 )x 2  8 abxy  (b 2  3a 2 ) y 2  0 with the line ax  by  c  0 form an equilateral triangle. (4) The area of a triangle formed by the lines ax 2  2hxy  by 2  0 and lx  my  n  0 is given by E3 n 2 h 2  ab am 2  2 hlm  bl 2 (5) The lines joining the origin to the points of intersection of line y  mx  c and the circle x 2  y 2  a 2 will be mutually perpendicular, if a 2 (m 2  1)  2c 2. equation of lines is (xy 1  yx 1 )2  d 2 (x 2  y 2 ) ID (6) If the distance of two lines passing through origin from the point ( x 1 , y 1 ) is d, then the ax 2  2hxy  by 2  2 gx  2 fy  c  0 will be (7) The lines represented by the equation equidistant from the origin, if f 4  g 4  c(bf 2  ag 2 ) U (8) The product of the perpendiculars drawn from ( x 1 , y 1 ) on the lines ax 2  2hxy  by 2  0 is given by ax 12  2 hx 1 y 1  by 12 D YG (a  b ) 2  4 h 2 (9) The product of the ax  2hxy  by 2  2 gx  2 fy  c  0 is 2 perpendiculars drawn from origin on the lines c (a  b)2  4 h 2 ST U (10) If the lines represented by the general equation ax 2  2hxy  by 2  2 gx  2 fy  c  0 are perpendicular, then the square of distance between the point of intersection and origin is f 2  g2 h2  b 2 (11) The square of distance between the point of intersection of the lines represented by c(a  b)  f 2  g 2 the equation ax 2  2hxy  by 2  2 gx  2 fy  c  0 and origin is ab  h 2 Example: 17 The area of the triangle formed by the lines 4 x 2  9 xy  9 y 2  0 and x  2 is (a) 2 Solution: (c) (b) 3 (c) 10 3 The area of triangle formed by the lines ax 2  2hxy  by 2  0 n 2 h 2  ab am 2  2hlm  bl 2 9 Here a  4 , b  9, h   , l  1, m  0, n  2 , then area of triangle 2 [Roorkee 2000] (d) and 20 3 lx  my  n  0 is given by Pair of Straight Lines 69 2 9 9 81 36 (2)2   4 4  2  2 4 2   30  10  = = 3 9 9  9  (1)2 The orthocentre of the triangle formed by the lines xy  0 and x  y  1 is Solution: (a) 1 1 (c)  ,  3 3 1 1 (d)  ,  4 4 Lines represented by xy  0 is x  0 , y  0. Then the triangle formed is right angled triangle at O(0, 0), therefore O(0, 0) is its orthocentre. Y x+y=1 O ID x=0 X y=0 If the pair of straight lines given by Ax 2  2 Hxy  By 2  0, (H 2  AB ) forms an equilateral triangle with U Example: 19 60 1 1 (b)  ,  2 2 (a) (0, 0) [IIT 1995] E3 Example: 18 line ax  by  c  0 then ( A  3 B)(3 A  B) is (b)  H Solution: (d) (c) 2H 2 D YG (a) H 2 [EAMCET 2003] (d) 4 H 2 We know that the pair of lines (a 2  3b 2 )x 2  8 abxy  (b 2  3a 2 )y 2  0 with the line ax  by  c  0 form an equilateral triangle. Hence comparing with Ax 2  2 Hxy  By 2  0 then 2H  8 ab ST U Now ( A  3 B)(3 A  B)  (8 a 2 )(8 b 2 )  (8 ab)2  (2 H )2  4 H 2. *** A  a 2  3b 2 , B  b 2  3a 2 ,