GROUP (A) – CLASS WORK PROBLEMSQ -3) Find the joint equation of the lines passing Q -1) Find the combined equation of the bisector through the point (–1, 2) and perpendicular to the lines x + 2y + 37 = 0 and of the angles between the co-ordinate axes. Ans. Let lines L1and L2be the bisector of the co- 3x – 4y – 53 = 0. Ans. Let lines L 1and L 2 be the lines passing ordinate axes θ1 and θ2be their inclinations through the point (–1, 2) and perpendicular to the lines x + 2y + 37 = 0 and 3x – 4y – 53 = π π and θ2 = – 4 4 L1 bisects first and 3rd quadrant θ1 = L2 bisects 2nd and 4th quadrant L2 Y 37 = 0 and 3x – 4y – 53 = 0 are – 1 and 2 –3 3 = respectively. –4 4 L1 π 4 π – 4 0 respectively. Slopes of the lines x + 2y + X ∴ Slopes of the lines L1and L2 are 2 and –4 3 respectively. Since the lines L1and L2 pass through the point (–1, 2), their equations Let slopes be m1 and m2 arey – 2 = 2 (x + 1) and y – 2 = –4 (x + 1) 3 m1 = tan π –π = 1 m2 = tan = –1 4 4 Since the lines are passing through origin ∴ y – 2 = 2x + 2 and 3y – 6 = – 4x – 4 ∴ 2 x – y + 4 = 0 4x + 3 y – 2 = 0 their equations are ∴ their joint equation is (2x – y + 4) (4x + 3y – 2) = 0 y = m1 x and y = m2 x ∴ 8x 2 + 6xy – 4x – 4xy – 3y2 + 2y + 16x + 12y –8=0 x – y = 0 and x + y = 0 Combined equation is (x – y ) (x + y ) = 0 ⇒ x 2 ∴ 8x 2 + 2xy – 3y2 + 12x + 14y – 8 = 0 2 –y =0 Q -2) Find the joint equation of lines which pass through the point (1, 2) and are parallel to the the lines co-ordinate axes. Ans. Equation of the coordinates axes are x = 0 and y = 0 ∴ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y = 2. i.e., x – 1 = 0 and y – 2 = 0 ∴ their joint equation is (x – 1) (y – 2) = 0 Q -4) Find the combined equation of lines through the origin, one of which is parallel and other is perpendicular to the line x + 2y +1857 = 0. Ans. The given equation of the line is x + 2y + 1857 = 0 Let slopes of one line be m1 and other line be m2 The slope line which is to the line x + 2y + 1857 = 0 is ∴ xy – 2x – y + 2 = 0 ∴ m1 = –1 2 –1 2 ∴ The slope of line which is ⊥ to the given line m2 = 2 ∴ The general equation of line passing through origin is y = mx Pair of Straight Lines Mahesh Tutorials Science 2 ∴ The equation of line to the general line is y = Q -4) Find the joint equation of lines passing through the origin having the inclinations 600 and 1200 –1 x 2 Ans. Slope of the line having inclination θ is x + 2y = 0 The equation of line ⊥ to the general line is y = 2x i.e. 2x – y = 0 ∴ The combined equation of line is (x + 2y) (2x – y) = 0 2x 2 + 4xy – xy – 2y2 = 0 2x 2 + 3xy – 2y2 = 0 tan θ. Inclinations of the given lines are 600 and 1200 ∴ their slopes are m1 = tan 600 = 3 and m2 = tan 1200 = tan(1800 – 600) = – tan 600 = – 3 . GROUP (A) – HOME WORK PROBLEMS Q -1) Find the joint equation of lines x – y = 0 and x + y = 0 (Ans. x2 – y2 = 0) Ans. The joint equation of the lines x – y = 0 and x + y = 0 is (x – y) (x + y) = 0 ∴ x2 – y2 = 0 y= and 2x – y – 1 = 0 (Ans. x2 + 3xy + y2 – 7x – 4y + 3 = 0) Ans. The joint equation of the lines x + y = 3 and 2x + y – 1 = 0 is (x + y – 3) (2x + y – 1) = 0 ∴ 2x2 + x y – x + 2x y + y2 – y – 6x – 3y + 3 = 0 ∴ 2x2 + 3x y + y2 – 7x – 4y + 3 = 0. Q -3) Find the joint equation of lines passing through the origin having slopes 3 and 2 Ans. We know that the equation of the line passing through the origin and having slope m is y = mx. Equation of the lines passing through the origin and having slope 3 and 2 are y = 3x and y = 2x respectively. i.e., their equations are 3x – y = 0 and 2x – y = 0 respectively. ∴ their joint equations is (3x – y)(2x – y) = 0 2 ∴ 6x – 3x y – 2x y + y = 0 ∴ 6x 2 – 5x y + y2 = 0. 3x and y = – i.e., 3x 3x – y = 0 and 3x + y = 0 ∴ the joint equation of these lines is ( Q -2) Find the joint equation of lines x + y = 3 2 Since the lines pass through the origin, their equations are 3x – y )( ) 3x + y = 0 ∴ 3x2 – y2 = 0. Q -5) Find the joint equation of lines which pass through the point (3,2) and are parallel to the lines x = 2 and y = 3 Ans. Equations of the lines are x = 2 and y = 3 ∴ the equations of the lines passing through (3, 2) and parallel to the lines x = 2 & y = 3 are x = 3 & y = 2. ∴ The joint equation is (x – 3)(y – 2) = 0 xy – 2x – 3y + 6 = 0 2x + 3y – xy – 6 = 0. Q -6) Find the joint equation of lines passing through the origin having slopes 1 + 3 and 1 – 3 Ans. The equation of the line passing through the origin and having slope m is y = mx Equations of the lines passing through the origin and having slopes 1+ 3 and 1 – 3 are ( ) ( ) y = 1 + 3 x and y = 1 – 3 x respectively i.e., The equation are (1 + 3 ) x – y = 0 ( ∴ The joint equation is Pair of Straight Lines ) and 1 – 3 x – y = 0 Mahesh Tutorials Science 3 1 – 3 x – y 1+ 3 x – y = 0 ( ) (x – 2 x + ( 3x – y 3x 2 )( x + ) ) 3x – y = 0 2 2 – xy – 3x – 3x + 3xy – xy – 3xy + y 2 = 0 –2x 2 – 2xy + y 2 = 0 ∴ equation of the line L2 is x = 3, i.e., x – 3 = 0 Hence, the equations of the required lines are x – 2y + 1 = 0 and x – 3 = 0 ∴ their joint equation is (x – 2y + 1)(x – 3) = 0 ∴ x 2– 2xy + x – 3x – 6y – 3 = 0 ∴ x 2– 2xy – 2x + 6y – 3 = 0. 2x 2 + 2xy – y 2 = 0 Q -7) Find the combined equation of lines which are parallel to the Y-axis and at the distance of 9 units from the Y-axis Ans. Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it are x = 9 and x = – 9 i.e., x – 9 = 0 and x + 9 = 0 Q -9) Find the combined equation of lines through the origin, which are perpendicular to the lines x + 2y = 9 and 3x + y = 8 Ans. Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y = 9 and 3x + y = 8 respectively. slope of given line are – Y 1 3 and – = – 3 respectively. 2 1 ∴ slope of the line L1 and L2 are 2 and O X′ 9 x = –9 1 respectively. 3 X 9 Since the lines L1 and L2 pass through the Y′ x=9 origin, their equations are ∴ their combined equation is (x – 9)(x + 9) = 0 ∴ x 2 – 81= 0. Q -8) Find the joint equation of lines passing through the point (3,2), one of which is parallel to the lines x – 2y = 29 and other one is perpendicular to the line y = 3 Ans. Let L1 be the line passes through (3,2) and parallel to the line x – 2y = 29 whose slope is 1 x 3 i.e., 2x – y = 0 and x – 3y = 0 ∴ their combined equation is (2x – y)(x – 3y) = 0 ∴ 2x2 – 6xy – xy + 3y2 = 0 ∴ 2x2 – 7xy + 3y2 = 0. Q-10) Find combined equation of lines through the origin, which are parallel to lines x + 3y = 27 and 2x = y – 3 Ans. Let L1 and L2 be the lines passing through the origin and parallel to the lines –1 1 = . –2 2 ∴ slope of the line L1 is 1 2 ∴ equation of the line L1 is y–2= y = 2x and y = 1 (x – 3) 2 ∴ 2y – 4 = x – 3 ∴ x – 2y + 1 = 0 Let L2 be the line passes through (3,2) and perpendicular to the line y = 3. ∴ equation of the line L2 is of the form x = a. Since L2 passes through (3,2), 3 = a x + 3y = 27 and 2x = y – 3 respectively. Slopes of the lines x + 3y = 27 and 2x = y – 3 are –1 –2 and 3 –1 ∴ slope of the lines L1 and L2 are –1 & 2 their 3 equations are y= –1 x & y = 2x 3 i.e., 3y + x = 0 & 2x – y = 0 ∴ The combined equation is Pair of Straight Lines Mahesh Tutorials Science 4 ∴The separate equations are (2x – y)(3y + x) = 0 2 2 6xy + 2x – 3y + xy = 0 2x 2 + 7xy – 3y2 = 0. ( ) ( ) y = 2 + 3 x and y = 2 – 3 x Q-11) Find the joint equation of lines which pass through (4, – 3) and are parallel to the lines x = 1 and y = 5 Ans. Equations of the lines are x = 1 and y = 5 ∴ the equations of the lines passing through (4, – 3) and parallel to the lines x = 1 and Q -3) Find the condition that the two lines of a2x2 + bcy2 = a (b + c)xy may be coincident and perpendicular Ans. Given equation is ⇒ a 2 x 2 + bcy 2 = a (b + c ) xy …(I) 2 2 Compare with Ax + 2Hxy + By = 0 y = 5 are x = 4 & y = – 3 i.e., x – 4 = 0 & y + 3 = 0 We get A = a 2 ; B = bc ; H = The joint equation is (x – 4)(y + 3) = 0 i) Equation will represent a pair of xy + 3x – 4y – 12 = 0 3x – 4y + xy – 12 = 0 coincident lines if H 2 – AB = 0 2 –a 2 2 (b + c ) – a .bc = 0 GROUP (B) – CLASS WORK PROBLEMS a2 2 (b + c ) – a 2bc = 0 4 Q -1) Find values of h, if lines given by 3x2 + 2hxy + 3y2 = 0 are real? ( (b (b ) a 2 b 2 + 2bc + c 2 – 4a 2bc = 0 Ans. Comparing the equation 3x2 + 2hxy + 3y2 = 0 with ax2 + 2hxy + by2 = 0, a2 we get, a = 3, h = h, b = 3. The lines represented by 3x2 + 2hxy + 3y2 = 0 a2 ) 2 + 2bc + c 2 – 4bc = 0 2 – 2bc + c 2 = 0 ) 2 are real, if h2 – ab ≥ 0 i.e., if h2 – 3(3) ≥ 0 i.e., if h2 ≥ 9 a 2 (b – c ) = 0 a (b – c ) = 0 Either a = 0 or b – c = 0 i.e., if h ≥ 3 or h ≤ – 3 Hence, the values of h are all real numbers Either a = 0 or b = c greater than or equal to 3 or less than or equal to – 3, i.e., (– ∞, 3] ∪ [3, ∞] = R – (– 3,3). lines if a = 0 or b = c Q -2) Find separate equation of the lines represented by x2 – 4xy + y2 = 0 Ans. Join equation x 2 – 4xy + y 2 = 0 2 2 i.e., x – 4xy + y = 0 y ∴ m 2 – 4m +1 = 0 ∵ x = m m= = 4± ( 4) 2 – 4 (1)(1) 2 4 ± 12 =2± 3 2 y ∴ = 2± 3 x ( ) ( ) ∴ y = 2± 3 x Pair of Straight Lines –a (b + c ) 2 = 4 ± 16 – 4 2 Equation represents two coincident ii) Equation (i) will represent a pair of ⊥ lines if A + B = 0 ⇒ a 2 + bc = 0 Q -4) Find the joint equation of a pair of lines through the origin and perpendicular to pair of lines 2x 2 – 3xy – 9y 2 = 0 Ans. Joint equation of lines is, 2x 2 – 3xy – 9y 2 = 0 So, Separate equation is, 2 – 9m 2 + 3m – 2 = 0 3y 9y 2 – 2 =0 x x y ∵ x = m 9m 2 + 6m – 3m – 2 = 0 3m ( 3m + 2) – 1 ( 3m + 2) = 0 Mahesh Tutorials Science 5 3m + 2 = 0 and 3m – 1 = 0 m= ∵ –2 1 m= 3 3 ⇒ k 2 = 36 ⇒ k = ± 6 y –2 y 1 = = x 3 x 3 Q -6) Find k, if one of the lines given by 2x + 3y = 0 …(i) – x + 3y = 0 …(ii) –2 Slope of equation (i) = 3 Slope of equation (ii) = 1 3 So, the slopes of ⊥ r Lines L1 and is, m1 = Ans. Equation of pair of lines, Equation of lines is, 2x + y = 0 y = –2x ∴ 3x + 2y = 0 y = 3 = –3x + y = 0 x Joint equation of perpendicular lines, ( 3x + 2y ) ( 3x – y ) = 0 9x 2 – 3xy + 6xy – 2y 2 = 0 2 9x + 3xy – 2y = 0 Q -5) Find k if the slope of one of the lines given by 5x 2 + kxy + y 2 = 0 is 5 times the slope of the other Ans. Given eq’n is 5x 2 + kxy + y 2 = 0 Solving (i) & (ii) Dividing by x2 –2kx 2 = 0 –2k = 0 k =0 Q -7) If one of the lines given by ax 2 + 2hxy + by 2 = 0 is perpendicular to the line x + my + n = 0 then show that al 2 + 2h m + bm 2 = 0 Ans. al 2 + 2h m + bm 2 = 0 is –l m r ∴ slope of ⊥ to lx + my + n = 0 is ax2 +2hxy + by2 = 0, a =5 ,2h = k, b = 1 ∴ Eqn.of line –2h –k = b 1 a 5 = b 1 ... (i) m1m2 = 5 ... (ii) But m1 = 5m2 ... (iii) Solving (i) & (iii) & Solving (i) & (ii) Passing through origin, m x ∴ m2 = –k 2 and m2 =1 6 ... (ii) Solving (i) & (ii) 2 m m ax 2 + 2hx + + b x = 0 x l ax 2 + 2hx 2 bm 2 x + 2 =0 l l ∴a + 2hm bm 2 + 2 =0 l l ∴ 5m2 + m2 = –k and 5m 2 .m 2 = 5 ∴ 6m2 = –k and 5m22 = 5 m l lx + my + n = 0 y= m1 + m 2 = –k ... (i) Slope of lx + my + n = 0 Comparing given eqn .with m1 m 2 = ... (ii) 4x 2 – 2kx 2 – 4x 2 =0 Lines L1 and L2 pass through the origin ∵ m1 + m 2 = ... (i) 4x 2 + kxy – y 2 = 0 2 Now, 2 4 x 2 + kxy – y 2 = 0 is 2x + y = 0 4x 2 + kx (–2x ) – ( –2x ) = 0 –3 m2 = 3 2 y –3 = x 2 2 k2 –k ⇒ = 1, =1 36 6 al 2 +2hlm + bm 2 =0 Pair of Straight Lines Mahesh Tutorials Science 6 Q -8) Find ‘α’if the sum of the squares of slopes of the lines repersented by (x + y = 0) αx – 3xy + y = 0 is 5. 2 2 ∴ Ans. Comparing the equation 2 = the distance of P (x, y) from the line x + 2y 1+ 4 2 ax – 3xy + y = 0 with ax2 + 2hxy + by2 = 0, ∴ we get, a = a, 2h = – 3, b = 1. Let m 1 and m 2 be the slopes of the lines represented by ax 2 – 3xy + y 2 = 0 ∴ m1 + m2 = – 2h = b and m1 + m 2 = – – ( –3 ) 1 x +y = ( x + 2y ) 1 +1 2 2 = 5 (x + y ) 2 2 ∴ 2 ( x + 2y ) = 5 ( x + y ) ( 2 ) ( ∴ 2 x 2 + 4xy + 4y 2 = 5 x 2 + 2xy + y 2 =3 ) ∴ 2x 2 + 8xy + 8y 2 = 5x 2 +10xy + 5y 2 ∴ 3x 2 + 2xy – 3y 2 = 0. a a = =a b 1 This is the required joint equation of the ∴ (m1 + m2)2 = (3)2 ∴ m12 + m22 + 2m1m2 = 9 lines which bisect the angle between the lines represented by x 2 + 3xy – 2y 2 = 0 ∴ m12 + m22 + 2a = 9 ∴ m12 + m22 = 9 – 2a Q-10 )S how But m12 + m22 = 5 ∴ 9 – 2a = 5 th at the lin es given by 3x – 8xy – 3y = 0 and x + 2y = 3 contain 2 2 the sides of an isosceles right angled triangle. ∴ 2a = 4 ∴ a = 2. Ans. Given equation 3x 2 – 8xy – 3y 2 = 0 Q-9) Find the joint equation of lines which bisect angles between lines represented by 3x 2 – 9xy + xy – 3y 2 = 0 3x ( x – 3y ) + y ( x – 3y ) = 0 x 2 + 3xy + 2y 2 = 0 ( x – 3y )( 3x + y ) = 0 Ans. x 2 + 3xy + 2y 2 = 0 ∴ x 2 + 2xy + xy + 2y 2 = 0 ∴ x (x + 2y) + y (x + 2y) = 0 ∴ (x + 2y) (x + y) = 0 ∴ s e pe rate e quation s ⇒ equations are ( x – 3y ) = 0 of the lin e s represented by x 2 + 2xy + xy + 2y 2 = 0 are x + 2y = 0 and x + y = 0. Angle bisector Y P(x,y) & ( 3x + y ) = 0 Slopes of these lines are m1 = 1 & m2 = – 3 3 Product of Slopes = m1m2 = 1 × – 3 = –1 3 ∴ Lines given by 3x 2 – 8xy – 3y 2 = 0 are ⊥ to each other. Slopes are m1 = …. (i) 1 and m2 = –3 3 X Slope of the line x + 2y = 3 is m = –1 2 Let θ1 be the angle between Let P (x, y) be any point on one angle bisector. x + 2y = 3 & x – 3y = 0 Since the points on the angle bisectors are equi - distant from both the lines, the From (i), (ii) & (iii) the given lines distance of P (x, y) from the line (x + 2y = 0) Pair of Straight Lines … (ii) form isosceles right angled triangle. Mahesh Tutorials Science Q-11 ) ∆ OAB is 7 formed by th e lines x – 4 xy + y = 0 and the line AB is given 2 2 by 2x + 3y – 1 = 0. Find the equation of the median passing through O. 3y = y= (x1,y1) Ans. 16 37 3y = 1 – 21 37 7 8 7 ∴D ≡ , 37 37 37 slope of OD, A D m= 7 / 37 8 / 37 m= 7 8 (x2,y2) B O equation of line OD is, y = mx Let A ≡ ( x1 , y1 ) , B ≡ ( x 2 , y2 ) Since, ≡ (x,y ) The combined equation of the lines passing through origin is x 2 – 4xy + y 2 = 0 The equation of line AB is, 1 – 2x 3 ...(iii) (1 – 2x ) 3 2 + (1 – 2x ) =0 9 2 9x 2 – 12x (1 – 2x ) + (1 – 2x ) = 0 9x 2 – 12x + 24x +1 – 4x + 4x 2 = 0 2 37x – 16x +1 = 0 Let, x1 and x 2 be the roots of the equation x1 + x 2 = the sides of an equilateral triangle. Find the area of the triangle. Ans. Let slope of OA be m and slope OB be 1 ∴ ∵ points A and B are intersecting point of lines passing through origin and line AB substituting (iii) in (i) x 2 – 4x Q-12) Show x2 + 4xy + y2 = 0 and x – y = 4 contain tan θ = 2x + 3y = 1 y= 7x 8 7x – 8y = 0 D is the mid-point of AB By mid-point formula x + x 2 y1 + y2 D≡ 1 , 2 2 y= 16 37 m1 – m2 1+ m1m2 3= m –1 1+m squaring ∴ 3 (1 + m)2 = m2 – 2m + 1 ∴ 3(m2 + 2m + 1) = m2 – 2m + 1 ∴ 3m2 + 6m + 3 = m2 – 2m + 1 ∴ 2m2 + 8m + 2 = 0 m2 + 4m + 1 = 0 2 ∴ y + 4 y +1 = 0 x x ... [By (i)] ∴ y2 + 4xy + x2 = 0 is the joint equation of lines passing through (0,0) and making 600 with x–y–4=0 x1 + x 2 8 = 2 37 A 8 y1 + y2 8 D≡ , ≡ 37 , y 37 2 Point D lies on the line 2x + 3y = 1 O M C 16 + 3y = 1 37 Pair of Straight Lines Mahesh Tutorials Science 8 GROUP (B) – HOME WORK PROBLEMS –4 OM = =2 2 2 Q -1) Determine the nature of lines represented P2 ∴ A(∆ ∆ OAB) = 3 8 = 3 sq units by following equations : i) x 2 + 2xy – y 2 = 0 Q-13) Show that the eqation x 2 – 16xy – 11y 2 = 0 ii) 4 x 2 + 4 xy + y 2 = 0 represents a pair of lines through the iii) x 2 + y 2 = 0 origin, each making an angle of 300 with iv) x 2 + 7xy + 2y 2 = 0 the line x + 2y – 1 = 0 v) px 2 + qy 2 = 0 , where p & q are real numbers. Ans. i) Comparing the equation x 2 + 2xy – y 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, 0 A 0 30 30 x + 2 y – 1= 0 B a = 1, 2h = 2, i.e., h = 1 and b = –1 ∴ h 2 – ab = 12 – 1 ( –1) = 1 +1 = 2 > 0 Ans. Let us find the joint equation OA and OB each makes an angle of 300 with x + 2y –1 = 0 Let slope of OA/OB be m1 (say) (m) ∴ slope of AB = m2 = – 1 2 m – m2 tan θ = 1 1+ m1m2 ∴ the lines represented by x 2 + 2xy – y 2 = 0 are real and distinct. ii) Comparing the equation 4x 2 + 4xy + y 2 = 0 with, ax 2 + 2hxy + by 2 = 0 we get, a = 4, 2h = 4, i.e. h = 2 and b = 1 2 ∴ ∴ 1 2 = 1 3 1+ – m 2 m+ 1 1 3 = 2m +1 2–m Squaring both sides 2 2 ( 2m +1) y = 2 x (2 – m ) ∴ (2 – m)2 = 3 (2m + 1)2 ∴ 4 – 4m + m2 = 3(4m2 + 4m + 1) ∴ 12m2 – m2 + 12m + 4m + 3 –1 = 0 ∴ 11m2 + 16m –1 = 0 Lines pass through origin ∴ y = mx is the equation is ∴h 2 – ab = ( 2) – 4 (1) = 4 – 4 = 0 ∴ the lines represented by 4x 2 + 4xy + y 2 = 0 are real and coincident. iii) Comparing the equation x 2 – y 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = 1, 2h = 0, i.e. h = 0 and b = –1 2 ∴ h 2 – ab = ( 0 ) – 1 ( –1) = 0 +1 = 1 > 0 ∴ the lines represented by x 2 – y 2 = 0 are real and distinct. iv) Comparing the equation x 2 + 7xy + 2y 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = 1, 2h = 7, b = 2, i.e. h = 2 ∴ 11 y +16 y – 1 = 0 x x 11y 2 y + 16 − 1 = 0 x x ∴ 11y2 + 16xy –x2 = 0 ∴ x2 – 16xy –11y2 = 0 is the required joint equation. ∴ 2 Pair of Straight Lines 7 2 2 49 7 ∴h 2 – ab = – (1)( 2) = –2 4 2 49 – 8 41 = = >0 4 4 ∴ The lines represented by x 2 + 7xy + 2y 2 = 0 are real and distinct. Mahesh Tutorials Science 9 v) Comparing the equation px 2 – qy 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = p, The auxilary equations is – m2 + 2m + 2 = 0 ∴ m2 – 2m – 2 = 0 2h = 0, i.e. h = 0 and b = –q 2 ∴ h 2 – ab = ( 0 ) – p ( –q ) = pq ∴ m= 2± ( –2) 2 = 2 ∴ the lines represented by px – qy = 0 are real and distinct. If p and q have different signs, then h 2 – ab = pq < 0 – 4 (1) ( –2 ) 2 ×1 If p and q have the same sign, then h 2 – ab = pq > 0 2 = 2± 4+8 2 2±2 3 =1 ± 3 2 ∴ m1 =1+ 3 and m2 =1 – 3 are the slopes of the lines. ∴ their separate equations are 2 2 ∴ the equation px – qy = 0 does not represent the lines. y = m1x and y = m2x ( ) ( ) i.e., y = 1 + 3 x and y = 1 – 3 x Q-2) Find separate equations of lines represented by following equations v) x 2 + 2xy cos ec α + y 2 = 0 2 y y + 2cos ec α + 1 = 0 x x i) 6x 2 – 5xy – 6y 2 = 0 ii) x 2 – 4y 2 = 0 iii) 3x 2 – y 2 = 0 Put iv) 2x 2 + 2xy – y 2 = 0 v) x + 2xy ( cos α ) + y = 0 2 y =m x (m )2 + 2 cos ec α (m ) + 1 = 0 2 Ans. i) 6x 2 – 5xy – 6y 2 = 0 m= −2 cos ec α ± 4 cos ec 2α − 4 × 1 × 1 2 m= −2 cos ec α ± 2 cot α 2 ∴ 6x2 – 9xy + 4xy – 6y2 = 0 ∴ 3x (2x – 3y) + 2y(2x – 3y) = 0 ∴ (2x – 3y)(3x + 2y) = 0 ∴ the separate equations of the lines are 2x – 3y = 0 and 3x + 2y = 0. m = − cos ec α ± cot α m= ii) x2 – 4y2 = 0 ∴ x2 – (2y)2 = 0 m= ∴ (x – 2y) (x + 2y) = 0 ∴ the separate equations of the lines are x – 2y = 0 and x + 2y = 0. 2 ( 3x ) ∴ ( 3x − y sin α y − (1 ± cos α ) = x sin α ∴ sin α y = − (1 + cos α ) x − y2 = 0 )( − (1 ± cos α ) y sin α = − (1 ± cos α ) x iii) 3x2 – y2 = 0 ∴ −1 cos α ± sin α sin α ) 3x + y = 0 ∴ the separate equations of the lines are 3x − y = 0 and & sin α y = − (1 − cos α ) x are the separate equations. 3x + y = 0 iv) 2x2 + 2xy2 – y2 = 0 Pair of Straight Lines Mahesh Tutorials Science 10 Q-3) If lines x2 + 2hxy + 4y2 = 0 are coincident m1 . m2 = find the values of h Ans. Comparing the equation a = –1 b The required lines are perpendicular to x2 + 2hxy + 4y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 1, h = h and b = 4. these lines Since the lines represented by x2 + 2hxy + 4y2 = 0 are coincident, h2 – ab = 0 ∴ The slopes are –1 –1 and m m1 2 These lines are passing through the ∴ h2 – 1 × 4 = 0 ∴ h2 = 4 origin, their separate equations are ∴ h = + 2. y= –1 –1 x and y = x m1 m2 Q-4) Find the combined equations of lines i.e., m1y = – x and m2y = – x through the origin which are perpendicular to the lines represented by x + m1y = 0 and x + m2y = 0 The combined equation is i) x2 + 4xy + 5y2 = 0 ii) x2 + xy – y2 = 0 (x + m1y) (x + m2y) = 0 x2 + (m1+ m2) xy + m1m2 y2 = 0 Ans. i) Comparing the equation x2 + 4xy + 5y2 = 0 with ax2 + 2hxy + by2 = 0, x2 + xy – y2 = 0 we get, a = 1, 2h = 4, b = 5 Let m1 and m2 be the slopes of the lines Q-5) Find k if sum of the slope of lines represented by the equation 3x2 + kxy + y2 = 0 is 0. represented by x2 + 4xy + 5y2 = 0 Ans. Comparing the equation 3x2 + kxy + y2 = 0 with ax2 + 2hxy + by2 = 0, –2h –4 = 5 b a 1 and m1m2 = = b 5 ∴ m1 + m2 = ... (i) represented by 3x2 + kxy + y2 = 0. Now required lines are perpendicular to these lines ∴ their slopes are – 1/m1 and – 1/m2 Since these lines are passing through the origin, their separate equations are –1 –1 y= x and y = x m1 m2 i.e., m1y = – x and m2y = – 0 ∴ their combined equations is (x + m1y) (x + m2y) = 0 ∴ x2 – 4 1 xy + y 2 = 0 5 5 ... [By (i)] ∴ 5x2 – 4xy + y2 = 0. ii) Comparing the equation x2 + xy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get, 1 a = 1, h = , b = – 1 2 Let m1 and m2 be the slopes of the lines represented by x2 + xy – y2 = 0 ∴ m1 + m2 = –2h –1 = =1 b –1 Pair of Straight Lines we get, a = 3, 2h = k, b = 1 Let m 1 and m 2 be the slopes of the lines ∴ m1 + m2 = –2h –k = = –k b 1 Now, m1 + m2 = 0 ∴–k=0 ... (Given) ∴ k = 0. Q-6) Find k if sum of the slope of lines represented by the equation 2x2 + kxy + 3y2 = 0 is 0 is equal to their product. Ans. Comparing the equation 2x2 + kxy + 3y2 = 0, with ax2 + 2hxy + by2 = 0, we get, a = 2, 2h = k, b = 3 Let m1 and m2 be the slopes of the lines represented by 2x2 + kxy + 3y2 = 0. ∴ m1 + m2 = –2h –k = b 3 and m1m2 = a 2 = b 3 Now, m1 + m2 = m1m2 ∴– k 2 = 3 3 ∴k=–2 ... (Given) Mahesh Tutorials Science 11 Q-7) Show that the difference between slope of lines repersented by the equation 3x2 + 4xy + y2 = 0 is 2. Ans. The given equation is 3x2 + 4xy + y2 = 0, ... (i) Comparing it with ax + 2hxy + by2 = 0, we get, a = 3, 2h = 4, b = 1 ∴ 45 3k – +3=0 25 5 ∴ 45 – 15k + 75 = 0 ∴ 15k = 120 ∴k=8 2 Let m1 and m2 be the slopes of lines given by(i) Then m1 + m 2 = – and m1m 2 = 2h –4 = = –4 b 1 a 3 = =3 b 1 ∴ (m1 – m2 )2 = (m1 + m2 )2 – 4m1m2 = (– 4)2 – 4(3) = 16 – 12 = 4 ∴ m1 – m2 = 2 ∴ the slopes of lines represented by 3x2 + 4xy + y2 = 0, differ by 2. Q-8) Find the value of k, if slope of one of the lines given by 3x2 – 4xy + ky2 = 0 is 1. Ans. The auxiliary equation of the lines given by 3x2 – 4xy + ky2 = 0 is km2 – 4m + 3 = 0. Q-10) The slope of one of the lines given by the equation 3x2 + 4xy + λy2 = 0 is 3 times the slope of the other line. Find λ. Ans. Comparing the equation 3x2 + 4xy + λy2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 3, 2h = 4, b = λ Let m1 and m2 be the slopes of the lines represented by 3x2 + 4xy + λy2 = 0. ∴ m1 + m2 = – and m1m 2 = 2h –4 = b λ a 3 = . b λ We are given that m2 = 3m1 ∴ m1 + 3m1 = –4 λ –4 λ ∴ m1 = – Given, slope of one of the lines is 1. ∴ 1 is the root of the auxiliary equation ∴ 4m1 = Let m1 and m2 be the slopes of lines given by km2 – 4m + 3 = 0. Also, m1 (3m1) = ∴ k(1)2 – 4(1) + 3 = 0 ∴k –4+3=0 ∴ m12 = ∴ k = 1. 1 1 ∴– = λ λ 1 λ ... (i) 3 λ 1 λ 2 Q-9) Find the value of k, if one of the lines given by 3x2 – kxy + 5y2 = 0 is perpendicular, to the line 5x + 3y = 0 Ans. Th e au xili ary equatio n of the lin es represented by ∴ ... [By (i)] 1 1 = λ2 λ ∴ λ = 1, as λ ≠ 0. 3x2 – kxy + 5y2 = 0 is 5m2 – km + 3 = 0. Q-11) If slopes of lines given by kx2 + 5xy + y2 = 0 differ by 1 then find the value of k. Now, one line is perpendicular to the line Ans. Comparing the equation kx2 + 5xy + y2 = 0 with ax2 + 2hxy + by2 = 0, 5x + 3y = 0, whose slope is ∴ slope of that line = m = ∴ –3 5 3 5 3 is the root of the auxiliary equation 5 we get, a = k, 2h = 5, b = 1. Let m1 and m2 be the slopes of the lines represented by kx2 + 5xy + y2 = 0 ∴ m1 + m2 = – 5m2 – km + 3 = 0. 2 3 3 ∴5 – k + 3 = 0 5 5 and m1m2 = 2h –5 = = –5 b 1 a k = =k b 1 Pair of Straight Lines Mahesh Tutorials Science 12 ∴ (m1 – m2)2 = (m1 + m2)2 – 4m1m2 = (– 5)2 – 4(k) = 25 – 4k ... (i) But m1 − m2 = 1 ∴ (m1 – m2)2 = 1 From (i) and (ii), 25 – 4k = 1 ∴ 4k = 24 (2x + y)(x + y) = 0 ∴ The se p arat e e quati on o f th e li ne s represented by 2x2 + 3xy + y2 = 0 are 2x + y = 0 and x + y = 0 ∴ k = 6. Y Angle bisector Q-12) Find the value of k, if one of the lines given by 6x2 + kxy + y2 = 0 is 2x + y = 0 P(x,y) Ans. Th e au xili ary equatio n of the lin es represented by 6x2 + kxy + y2 = 0 is m2 + km + 6 = 0. Since one of the line is 2x + y = 0 whose slope is m = – 2 ∴ – 2 is the root of the auxiliary equation m2 + km + 6 = 0. ∴ (– 2)2 + k (– 2) + 6 = 0 ∴ 4 – 2k + 6 = 0 Angle X bisector Let P (x, y) be any point on one angle bisector. Since the points on the angle bisectors are equidistant from both the lines, the distance of P (x, y) from the line 2x + y = 0 ∴ the distance of P (x, y) from the line x + y =0 ∴ 2k = 10 ∴ k = 5. ∴ 2x + y 4 +1 x +y = 1 +1 Q-13) If the sum of the slopes of the lines given by 2x2 + kxy – 9y2 = 0 is equal to five times their product, find the value of k. Ans. Comparing the equation 2x2 + kxy – 9y2 = 0, with ax2 + 2hxy + by2 = 0, we get, a = 2, h = represented by 2x2 + kxy – 9y2 = 0 k –2 2h 2 = k ∴ m1 + m2 = – = b –9 9 5 ( x + y )2 = 2 2 ∴ 2 ( 2x + y ) = 5 ( x + y ) 2 ∴ 2 4x 2 + 4xy + y 2 = 5 x 2 + 2xy + y 2 3x 2 – 2xy – 3y 2 = 0 This is the required joint equation of the lines which bisect the angle between the lines represented by 2x 2 + 3xy + y 2 = 0 Q-15) Find the joint equation of the pair of lines through origin and making an equilateral triangle with the line x = 3 a –2 = b 9 Now, m1 + m2 = 5m1m2 k –2 = 5 ; 9 9 ( 2x + y )2 8x 2 + 8xy + 2y 2 = 5x 2 +10xy + 5y 2 k ,b=–9 2 Let m1 and m2 be the slopes of the lines m1m2 = ∴ ... (given) Ans. Let OA and OB be the lines through the origin making an angle of 600 with the line x = 3. k = –10 Y Q-14) Find the joint equation of lines which bisect angles betweeen lines represented by 2x2 + 3xy + y2 = 0 Ans. 2x2 + 3xy + y2 = 0 A 1500 600 30 0 X' O 30 X 0 600 2x2 + 2xy + xy + y2 = 0 2x (x + y) + y (x + y) = 0 B x=3 Pair of Straight Lines Y' Mahesh Tutorials Science 13 ∴ OA and OB make an angle of 300 and 1500 with the positive direction of X-axis ∴ slope of OA = tan 300 = 1 1 3 x Slope of OB = tan1500 = tan (1800 – 300) = – tan300 = –1 3 ∴ equation of the line OB is y = ∴ 3y = – x 3y –1 3 x ∴ x + 3y = 0 )( x + ) 3y = 0 i.e., x2 – 3y2 = 0. Q-16) Find k, if the sum of the slopes of the lines represented by the equation x2 + kxy – 3y2 = 0 is twice their product Ans. Given equation is x2 + kxy – 3y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0 we get, a = 1 , 2h = k , b = –3 Let m 1 and m 2 be the slopes of the lines represented by x2 + kxy – 3y2 = 0 ∴ m1 + m2 = –2h k = b 3 a 1 and m1m2 = = – 3 b According to the given condition. m1m2 = 2 (m1m2) ∴ k 1 = 2 − 3 2 ...(given) ∴ k=4 ∴ k=±2 kx2 + 4xy – y2 = 0 exceeds the slops of the other by 8, find k Ans. Given equation is kx2 + 4xy – y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0 we get, a = k , 2h = 4 , b = –1 Let m 1 and m 2 be the slopes of the lines represented by kx2 + 4xy – y2 = 0 ∴ m1 + m2 = 4 and m1m2 = –k According to the given condition m2 = m1 + 8 ∴ m1 + (m1 + 8) = 4 ∴ 2m1 = – 4 ∴ m1 = – 2 Now, m1 (m1 + 8 ) = –k (– 2) (–2 + 8 ) = –k ∴ (–2) (6) = –k ∴ –12 = –k ∴ k = 12 Q-19) Find the condition that the line 3x +y = 0 may be perpendicular to one of the lines represented by ax2 + 2 hxy + by2 = 0 Ans. Th e au xili ary e quatio n of the lin e s represented by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0 Since one line is perpendicular to the line 3x + y = 0 whose slope is – ∴ k=–3 Q-17) Find k , if th e s lo pes of th e lines represented by the equation 3x2 + kxy – y2 = 0 differ by 4 Ans. Given equation 3x2 + kxy – y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0 we get, a = 3 , 2h = k , b = –1 Let m 1 and m 2 be the slopes of the lines represented by 3x2 + kxy – y2 = 0 ∴ m1 + m2 = But |m1 – m2| = 4 ∴ (m1 – m2)2 = 16 Q-18) If the slope of one of the lines given by ∴ required combined equation of the lines is (x – ...(i) ∴ From (i) and (ii), we get k2 + 12 = 16 ∴ x – 3y = 0 3y = x = k2 – 4(–3) = k2 + 12 3 ∴ equation of the line OA is y = ∴ Now, (m1 – m2)2 = (m1 + m2)2 – 4m1m2 a –2h = k and m1 m2 = = –3 b b ∴ slope of that line = m = ∴ m= 3 = –3 1 1 3 1 is the root of the auxiliary equation 3 bm2 + 2hm + a = 0 2 1 1 ∴ b + 2h + a = 0 3 3 ∴ b + 6h + 9a = 0 ∴ 9a + 6h + b = 0 This is the required condition Pair of Straight Lines Mahesh Tutorials Science 14 Q-20) Show that 3x + 4y + 5 = 0 and the lines (3x +4y )2 –3 (4x +3y)2 = 0 form the side an equilateral triangle. Q-21) Show that x2 – 4xy + y2 = 0 and the lines x + y – 6 = 0 form an equilateral triangle Find its area and perimeter –3 Ans. The slope of the line 3x + 4y + 5 = 0 is m1 = 4 Let m be the slope of one of the line making an angle of 600 with the line 3x + 4y + 5 = 0 Ans. Le t O A and OB be Since the angle between the lines having slope m and m1 is 600 origin Let M be tan 600 = m – m1 1+ m.m1 Y B 0 60 the lines th ro u g h M x +y = 60 the slope of OA or 0 A O X OB which –3 m– 4 3 = –3 1+ m 4 from equilateral triangle with line x + y = 6 whose slope is ∴ Its equation is y = mx tan 600 = 4m + 3 3 = 4 – 3m Squaring both sides, we get 2 ( 4m + 3) 3= 2 ( 4m – 3) 3(4 – 3m)2 = (4m + 3)2 ; ∴ 3 (16 – 24m + 9m2) = 16m2 + 24m + 9 ∴ 48 – 72m + 27m2 = 16m2 + 24m + 9 ∴ 11m2 – 96m + 39 = 0 Since, required lines pass through the origin, their equations is of the form y = mx i.e. m = 6 y x 2 y y 11 – 96 + 39 = 0 x x 2 3= m +1 ; 1–m ...(i) 3= m +1 1–m m 2 + 2m +1 m 2 – 2m +1 ∴ 3m2 – 6m + 3 = m2 + 2m + 1 ∴ 2m2 – 8m – 2 = 0 ∴ m2 – 4m – 1 = 0 Joint equation of the pair of lines is given by y y2 –4 +3 =0 x x2 ...[From (i)] x2 – 4xy + y2 = 0 which is given in question He nce, joint equation from e quilateral triangle with x + y = 6 ⊥ distance of AB from origin is OM = y 96y 11 2 – + 39 = 0 x x – 6 1+1 = 3 ( OM) 2 ( 3) = 11y2 – 96xy + 39x2 = 0 39x2 – 96xy + 11y2 = 0 ∴ Area of ∆ OAB = is the required joint equation which can be written as 39x2 – 96xy + 11y2 = 0 i.e. (9x2 – 48x2) + (24xy + 72xy) + (16y2 – 27y2) = 0 In right – angled triangle OAM i.e. (9x2 + 24xy + 16y2) – (48x2 – 72xy + 27y2) = 0 i.e. (9x2 + 24xy + 16y2) –3(16x2 – 24xy + 9y2) = 0 i.e. (3x + 4y)2 –3(4x – 3y) = 0 Hence, the line 3x + 4y + 5 = 0 and the lines (3x + 4y)2 –3(4x – 3y2) from an equilateral triangle. Pair of Straight Lines 3 2 3 = 3 sq units. sin 600 = OM OA 3 3 = ∴ OA = 2 2 OA ∴ Length of the each side of the equilateral triangle OAB = 2 units ∴ perimeter of ∆ OAB = 3 × length of each side = 3 × 2 = 6 units ∴ Mahesh Tutorials Science 15 Q-22) Find the joint equation of lines through the origin, each of which makes angle of 30 with the line 3x + 2y + 66 = 0 Q -1) Find the acute angle between the lines given by x2 + y2 = 2xycosec α Ans. Comparing the equation Ans. The slope of the line3x + 2y + 66 = 0 is m1 = – GROUP (C) – CLASS WORK PROBLEMS 0 3 2 Let m be the slope of one of the lines making x2 +y2 = 2xy cossec α with ax2 + by2 + 2hxy = 0 we get a = 1 , b = 1 , h = – cosecα an angle of 300 with the 3x + 2y + 66 = 0. The angle between the lines having slopes m and Let θ be the acute anle between the lines m1 is 300 tan θ = ∴ tan 300 = m – m1 1 , where tan 300 = 1+ m.m1 3 3 m – – 2 ∴ = 3 1+ m – 3 2 1 ∴ 1 3 = 2m + 3 2 – 3m tan θ = tan θ = 2 h 2 – ab a +b 2 cosec 2 α – 1 2 cot 2 α tan θ = cot α π tan θ = tan – α 2 On squaring both the sides, we get, 2 1 ( 2m + 3) = 3 ( 2 – 3m )2 θ = ∴ (2 – 3m)2 = 3(2m + 3)2 ∴ 4 – 12m + 9m2 = 3(4m2 + 12m + 9) ∴ 4 – 12m + 9m2 = 12m2 + 36m + 27 Q -2) F in d th e acute an gle betw een th e linesgiven by ∴ 3m2 + 48m + 23 = 0 This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = π –α 2 y x (a2 – 3b2) x2 + 8abxy + (b2 – 3a2)y2 = 0 Ans. Combined equation of lines is, (a 2 ) ( ) – 3b 2 x 2 + 8abxy + b 2 – 3a 2 y 2 = 0 comparing with, Ax 2 + 2Hxy + By 2 = 0 So, ∴ the joint equation of the two lines is A = a 2 – 3b 2 ; H = 4a ; B = b 2 – 3a 2 2 y y 3 + 48 + 23 = 0 x x ∴ tan θ = 3y 2 48y + + 23 = 0 x x2 2 2 2 2 ∴ 3y + 48xy + 23x = 0; ∴ 23x + 48xy + 3y = 0. tan θ = 2 h 2 – ab a +b ( )( 2 16a 2b 2 – a 2 – 3b 2 b 2 – 3a 2 2 2 2 a – 3b + b – 3a ) 2 Q-23) Find the condition that the line 4x +5y = 0 coincides with one of the lines given by ax2 + 2hxy + by2 = 0 tan θ = 2 16a 2b 2 – 10a 2b 2 + 3a 4 + 3b 4 ( –2 a 2 + b 2 ) Ans. ax2 + 2hxy + by2 = 0. 2 y y ax + 2h + b = 0 x x tan θ = 2 3a 4 + 6a 2b 2 + 3b 4 ( –2 a 2 + b 2 4x +5y = 0 ; 5y = –4x 2 –4 y –4 –4 = ; a + 2h + b =0 5 x 5 5 a– 8h 16b – = 0 ∴ 25a – 40h + 16b = 0 25 5 tan θ = ( – 3 a2 + b2 θ = tan –1 (a 2 +b ( 3) ; 2 ) ) ; tan θ = 3 ) θ= π 3 Pair of Straight Lines Mahesh Tutorials Science 16 Q -3) If the angle between the lines ax2 + 2hxy + by2 = 0 is equal to the angle between the lines 2x2 – 5xy + 3y2 = 0 prove 2 2 tan 45 = 2 that 100(h – ab) = (a + b) Ans. 1st combined equation is, ax 2 + 2hxy + by 2 = 0 ... (i) h 2 – 24 5=2 4 So, tan θ = h2 –6 4 5 5 = h 2 – 24 2 h 2 – ab a +b h 2 – 24 = 25 2nd combined equation is, h 2 = 25 + 24 2x 2 – 5xy + 3y 2 = 0 h 2 = 49 A = 2, H = h = ±7 –5 , B =3 2 Q -5) Show that one of the straight lines given by ax2 + 2hxy + by2 = 0 bisects an angle 25 2 –6 4 tan θ = 5 between the co-ordinate axes if (a + b)2 = 4h2 Ans. The equation of a straight line is given by, ax 2 + 2hxy + by 2 = 0 1 2 4 tan θ = 5 tan θ = Divide by x 2 , 2 y y a + 2h + b = 0 x x put y = mx 1 5 As per given, The angle between these two lines is equal ∴ 2 h 2 – ab 1 = a +b 5 ( ) a + 2hm + bm 2 = 0 ∴ bm 2 + 2hm + a = 0 As the lines bisects an angle between coordinate axes. 10 h 2 – ab = a + b 100 h 2 – ab = (a + b ) y x ∴m = 2 So, ... Hence proved Q -4) If acute angle between 3x2 + hxy + 2y2 = 0 is of measure 450, find h. Ans. The combined equation is, θ = 450 ∴ m = tan θ ∴ m = tan 450 ∴m =1 2 3x 2 + hxy + 2y 2 = 0 ∴ b (1) + 2h (1) + a = 0 comparing with, b + 2h + a = 0 Ax 2 + 2Hxy + By 2 = 0 (a + b ) = –2h h A = 3, H = , B = 2 2 tan θ = 2 h 2 – ab a +b Pair of Straight Lines squaring on both sides, (a + b ) 2 = 4h 2 Mahesh Tutorials Science 17 GROUP (C) – HOME WORK PROBLEMS Q -1) Find the measure of the acute angle between the lines represented by : i) 3x 2 – 4 3xy + 3y 2 = 0 2 2 ∴ tan θ = 3 5 3 ∴ θ = tan–1 . 5 iii) Comparing the equation ii) 4 x + 5xy + y = 0 2x 2 + 7xy + 3y 2 = 0 with iii) 2x 2 + 7xy + 3y 2 = 0 ax 2 + 2hxy + by 2 = 0 , we get, Ans. i) Comparing the equation 7 ,b=3 2 3x 2 – 4 3xy + 3y 2 = 0 with a = 2, h = ax 2 + 2hxy + by 2 = 0 , we get, Let θ be the acute angle between the lines. a = 3, 2h = –4 3 , i.e., h = –2 3 and b = 3 Let θ be the acute angle between the lines. ∴ tan θ = 2 h 2 – ab a +b ∴ tan θ = 2 = ( –2 3 ) 2 2 7 2 – ( 2)( 3 ) 2 = 2+3 – 3 (3) 3+3 2 = 2 h 2 – ab a +b 2 12 – 9 2 3 = 6 6 1 ∴ tan θ = 3 = = tan 300 = ∴ θ = 300 ii) Comparing the equation 4x 2 + 5xy + y 2 = 0 with = a = 4, 2h = 5, i.e., h = 5 and b = 1. 2 Let θ be the acute angle between the lines. 2 h 2 – ab a +b ∴ tan θ = 2 5 2 – 4 (1) 2 = 4 +1 2 49 – 24 5×2 25 5 = ax 2 + 2hxy + by 2 = 0 , we get, 49 –6 4 5 5 5 ∴ tan θ = 1 θ = 450 Q -2) Find the value of b, if the angle between the lines given by 6x2 + xy + by2 = 0 is 450 Ans. Comparing the equation 6x2 + xy + by2 = 0 with Ax 2 + 2Hxy + By 2 = 0 , we get, A = 6, 2H = 1, i.e., H = 25 3 2 –4 2× 4 2 = = 5 5 1 and B = b 2 Since the angle between the lines is 450, tan 450 = 2 H 2 – AB A +B Pair of Straight Lines Mahesh Tutorials Science 18 Q -4) If the lines given by ax2 + 2hxy + by2 = 0 2 1 2 – 6 (b ) 2 ∴1 = 6 +b 2 ∴1 = form an equilateral triangle with the line lx + my = 1, show that (3a + b) (a +3b) –4h2 = 0 Ans. Since the lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines 1 – 6b 4 6+b ax 2 + 2hxy + by2 = 0 is 600 2 h 2 – ab a +b ∴ tan 600 = 1 ∴ (6 + b)2 = 4 – 6b 4 2 ∴ 36 + 12b + b = 1 – 24b ∴ b2 + 36b + 35 = 0 ∴ b2 + 35b + b + 35 = 0 ∴ b (b + 35) + 1(b + 35) = 0 ∴ (b + 1) + 1(b + 35) = 0 ∴ b + 1 = 0 or b + 35 = 0 ∴b=–1 or b = – 35. ∴ ∴ 3a2 + 10ab + 3b2 = 4h2 ∴ 3a2 + 9ab + ab + 3b2 = 4h2 a2 – 12b2)x2 + 16abxy + (4b2 – 3a2)y2 = 0 is 600 ( Ans. Given combined equation is, ) ( ) a 2 – 12b 2 x 2 +16abxy + 4b 2 – 3a 2 y 2 = 0 comparing with, Ax 2 + 2Hxy + By 2 = 0 A = a 2 – 12b 2 B = 4b 2 – 3a 2 = 2 64a 2b 2 – 40a 2b 2 + 3a 2 + 48b 2 a 2 – 12b 2 + 4b 2 – 3a 2 = 2 24a 2b 2 + 3a 2 + b 2 + 48b 4 – 8b 2 – 2a 2 ( 2 3 4b 2 + a 2 ( 2 –2 4b + a ∴ 3a(a + 3b) + b(a + 3b)= 4h2 ∴ (3a + b) (a + 3b) – 4h2 =0 This is the required condition Q-5) Find the value of k, if the lines x2 + kxy + y2 = 0 and x + y = 1 contain the sides of an equilateral triangle. Also, find the area of the triangle. Ans. Since the lines x2 + kxy + y2 = 0 form an equilateral triangle with the line x + y = 1, the angle between the lines H = 8ab = 2 h 2 – ab a +b ∴ 3(a +b)2 = 4 (h2 – ab) ∴ 3(a2 + 2ab + b2) = 4h2 – 4ab ∴ 3a2 + 6ab + 3b2 + 4ab = 4h2 Q -3) Show that the angle between ( 3= 2 x2 + kxy + y2 = 0 is 600 Comparing the equation x2 + kxy + y2 = 0 with ax2 + 2hxy + by2 = 0 a = 1, 2h = k, b = 1 tan 60 = ) ) 2 h 2 – ab a +b 2 tan 60 = (k 2) 2 – (1)(1) 1 +1 = – 3 3= tan θ = 3 θ = tan –1 2 k2 4 –1 2 ( 3) θ = 60 ... Hence proved 3= k2 – 4 2 On squaring both sides 12 = k2 – 4 k2 = 16 Pair of Straight Lines Mahesh Tutorials Science 19 k = ±4 Let the length of the perpendicular from the given to the line x + y = 1 be ‘p’ Area of the equilateral triangle = = p2 3 sq units = 1 2 3 1 1 3 2 sq units Q -6) Find the value of b, if θ is the measure of angle between the line 3x2 + 4xy + by2 = 0 and tan θ = with Ax 2 + 2Hxy + By2 = 0 we get A = 3, 2H = 4 i.e. H = 2 B = b tan θ = 1 2 [gn] 2 H 2 – AB A+B 1 = 2 2 22 – 3b 3+b 1 = 2 2 4 – 3b 3 +b On squaring both sides 1 = 4 4 ( 4 – 3b ) (3 + b ) get, x1 = – 1/4 and y1 = 1/2 ∴ the point of intersection is P(– 1/4, 1/2) Q -2) Find combined equation of a pair of lines through the point (2, -3) and perpendicular to the lines given by x 2 + xy – 2y 2= 0. Ans. x 2 + xy – 2y 2 = 0 x 2 + 2xy – xy – 2y 2 = 0 x ( x + 2y ) – y ( x + 2y ) = 0 ( x – y )( x + 2y ) = 0 1 2 Ans. Comparing the equation 3x 2 + 4xy + by2 = 0 Now, tanθ = ∴ 2x1 + y1 = 0 and 2x1 – y1 + 1 = 0 Solving these equations simultaneously, we 2 9 + 6b + b2 = 64 – 48b b2 + 54b – 55 = 0 b2 + 55b – b – 55 = 0 (b + 55) (b – 1) = 0 x – y = 0 or x + 2y = 0 Slope of x – y = 0 is 1. Slope of line ⊥r to x – y = 0 is –1 ∴ equation of line ⊥r to x – y = 0 and passing through ( 2, –3 ) is, y + 3 = –1 ( x – 2) y + 3 = –x + 2 x + y +1 = 0 ... (i) Slope of x + 2y = 0 is –1 2 ∴ Slope of line ⊥r to x + 2y = 0 is 2 ∴ equation of line ⊥r to x + 2y = 0 and passing through ( 2, –3 ) is, y + 3 = 2 ( x – 2) y + 3 = 2x – 4 2x – y – 7 = 0 combined equations of lines is, ... (ii) ( x + y +1)( 2x – y – 7 ) = 0 from (i) and (ii) ∴ b = –55 & b = 1 2 GROUP (D) – CLASS WORK PROBLEMS 2x – xy – 7x + 2xy – y 2 – 7y + 2x – y – 7 = 0 2x 2 + xy – y 2 – 5x – 8y – 7 = 0 Q -1) Find the separate equation of the lines represented by 4x2 – y2 + 2x – y = 0. Also find the point of intersection of these lines. Ans. The combined equation of the lines is 4x 2 – y2 + 2x – y = 0 ∴ (2x + y)(2x – y) + (2x + y) = 0 ∴ (2x + y)(2x – y + 1) = 0 ∴ their separate equations are 2x + y = 0 and 2x – y + 1 = 0 Let P (x1,y1) be their point of intersection. Then (x1, y1) satisfies these equations Q -3) Find k if kx 2 – xy + 2y 2 + 19x – 6y – 20 = 0 represents pairs of lines. Further find wh ether th ese lines are parallel or intersecting Ans. Given equation is kx 2 – xy + 2y 2 +19x – 6y – 20 = 0 Comparing with ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 Pair of Straight Lines Mahesh Tutorials Science 20 we get a = k,h = –1 19 , b = 2, g = , f = –3, c = –20 2 2 ∴ Equation represents a pair of lines abc + 2 fgh – af 2 2 2 – bg – ch = 0 57 361 – +5 = 0 2 2 –49k – 304 +5 =0 2 px 2 – 8xy + 3y2 + 14x +2y + q = 0 Comparing it with ax 2 + 2hxy + by2 +2gx + 2fy + c = 0, we get a = p, h = –4 , b = 3, g = 7 f = 1 and c = q 19 1 × –k ×9 –2 2 2 361 1 × + 20 × = 0 4 4 –40k + ( –6 ) × –49k + Q -5) Find p and q, if the equation px2 – 8xy + 3y2 + 14x +2y + q = 0 represents a pair of perpendicular lines Ans. Given equation is Since the given equation represents a pair of line perpendicular to each other, ∴ a+b=0 ∴ p+3=0 ∴ p = –3 Also the equation represents a pair of straight lines –49k – 152 + 5 = 0 a h ∴ g –49k = 147 k = –3 2 –1 Consider h 2 – ab = – k × 2 2 = ∴ lines are intersecting (x –1)2 – (y –1)2 – (x –1) – 5 (y –1) + λ = 0. represents two straight lines. Ans. Given equation is (x –1)2 – (y –1)2 – (x –1) – 5 (y –1) + λ = 0. ...(i) ) ( ) – 2x +1 – y 2 – 2y +1 –x + 1 –5y + 5 +λ λ= 0 x2 –2x + 1 – y2 +2y – 1 –x + 1 –5y + 5 +λ λ=0 2 2 x –3x – y –3y + 6 +λ λ=0 x2 – y2 –3x –3y + 6 +λ λ=0 Comparing with ...(ii) –3 –3 ,f= , c = 6 +λ λ 2 2 ∴ Equation (ii) represents a pair of lines 2 2 abc + 2fgh – af – bg 2 – ch = 0 1 × –1 × (6 + λ ) + 0 –1 × 9 9 –6 + λ – + =0 4 4 –6 – λ = 0 λ = –6 3 1 =0 7 1 q ∴ ∴ ∴ ∴ –25q – 200 = 0 – 25(q + 8) = 0 q = –8 p = –3 and q = –8 Q -5) Show that one of the straight lines given by ax2 + 2hxy + by2 = 0 bisects an angle between the co-ordinate axes if (a + b)2 = 4h2 Ans. The equation of a straight line is given by, ax 2 + 2hxy + by 2 = 0 Divide by x 2 , ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 a = 1, h = 0, b = –1, g = ∴ –4 ∴ – 3 (3q – 1) + 4 (–4q – 7) + 7 (–4 – 21) = 0 ∴ –9q + 3 – 16q – 28 – 175 = 0 Q -4) Find λ so that equation 2 9 9 +1× –0=0 4 4 2 y y a + 2h + b = 0 x x put y = mx ∴m = y x a + 2hm + bm 2 = 0 ∴ bm 2 + 2hm + a = 0 As the lines bisects an angle between coordinate axes. So, θ = 450 Pair of Straight Lines g f =0 c –3 –4 7 1 25 +6= ≠0 4 4 (x h b f Mahesh Tutorials Science 21 ∴ m = tan θ tanθ = 1 ∴ m = tan 450 θ= ∴m =1 2 ∴ b (1) + 2h (1) + a = 0 π 4 GROUP (D) – HOME WORK PROBLEMS b + 2h + a = 0 (a + b ) = –2h squaring on both sides, 2 (a + b ) = 4h 2 Q -6) Show that 2x2 + 7xy +3y2 – 5x – 5y + 2 = 0 represents a pair of lines and find the acute angle between them. Ans. Comparing 2x2 + 7xy +3y2 – 5x – 5y + 2 = 0 with ax2 + 2hxy + by2 +2gx + 2fy + c = 0 we get a = 2 ,h = 7 –5 –5 , b = 3, g = , f= and c = 2 2 2 2 Q -1) Find the combined equation if the pair of lines through (2,3) one of which is parallel to 2x + 3y = 5 and other perpendicular to x – 4y =7. Ans. Let L1 be the line throught (2,3) and parallel to the line 2x + 3y = 5 The slope of the line 2x + 3y = 5 is – ∴ the slope of the line L1is – 2 3 2 and it passes 3 through (2,3) ∴ equation of the line L1is y – 3 = – 2 (x – 2) 3 condition for pair of lines is ∆ = 0 LHS = abc + 2fgh – af 2 – bg2 – ch2 ∴ 3y – 9 = – 2x + 4 ∴ 2x + 3y – 13 = 0 –5 = 2 × 3 × 2 + (–5) 2 7 2 Le t L 2 be the line through (2,3) and perpendicular to the line x – 4y = 7. 25 25 49 –3 × –2× 4 4 4 –1 1 = . –4 4 ∴ the slope of the line L2 is – 4 and it passes through (2,3) –2× = 12 + 25 × 7 50 75 98 – – – 4 4 4 4 = 12 + 175 – 125 98 = 4 4 50 98 = 12 + – 4 4 48 = 12 – 4 = 12 – 12 =0 ∴ equation of the line L2 is y – 3 = – 4(x – 2) ∴ y – 3 = – 4x + 8 ∴ 4x + y – 11= 0 Hence, the equations of required lines are 2x + 3y – 13 = 0 and 4x + y – 11= 0. ∴ their combined equation is (2x + 3y – 13) (4x + y – 11) = 0. ∴ The equation represents a pair of straight lines If θ is the acute angle between the lines then tan θ = The slope of the line x – 4y = 7 is 2 h 2 – ab a +b 2 7 49 2 – 2 (3) 2 –6 2 4 = = 2+3 5 ∴ 8x 2 + 2xy – 22x + 12xy + 3y2 – 33y – 52x – 13y + 143 = 0 ∴ 8x 2 + 14xy + 3y2 – 74x – 46y + 143 = 0. Q -2) Find the joint equation of the pair of lines through (– 2, 2), one of which is parallel to the linex + 2y – 3 = 0 and other is perpendicular to the line y = 3 Ans. Let L1 be the line passes throught (– 2,2) and parallel to the line x + 2y – 3 = 0 whose slope is –1 2 Pair of Straight Lines Mahesh Tutorials Science 22 ∴ Slope of line L1 is Q -4) Find k if x 2 – 3xy + 2y 2 + x – y + k = 0 –1 2 represents a pair of lines Ans. x 2 + 3xy + 2y2 + x – y + k = 0 ∴ equation of the line L1 is (y – 2) = –1 ( x + 2) 2 a = 1, b = 2, h = 2y – 4 = – x – 2 Condition for pair of lines. abc + 2fgh – af 2 – bg 2 – ch2 = 0 x + 2 + 2y – 4 = 0 x + 2y – 2 = 0 Let L2 be the line passes throught (– 2, 2) and perpendicular to the line y = 3 ∴ equation of the line L2 is of the form x = a Since L2 passes throug (– 2, 2), – 2 = a ∴ equation of the line L2 is x = – 2 i.e., x + 2 = 0 Hence the equation of the required line are x + 2y – 2 = 0 and x + 2 = 0 ∴ The joint equation is (x + 2y – 2) (x + 2) = 0 1 1 9 3 2k + 2 – 1 – 2 – k = 0 4 4 4 8 ∴ 2k – 3 1 1 9k – – – =0 4 4 2 4 8k – 3 – 1 – 2 – 9k = 0 –k = 6 ∴ k =–6 Q -5) Find the separate equations for each of the following lines 10(x + 1)2 + (x + 1) (y – 2) – 3 (y – 2)2 = 0 Ans. The given equation is 2 x + 2x + 2xy + 4y – 2x – 4 = 0 x 2 + 2xy + 4y – 4 = 0 10(x +1) + (x +1) (y – 2) – 3 (y – 2) = 0 ...(i) Let x + 1 = X and y – 2 = Y Q -3) Show that the equation ∴ The given equation becomes 10X2 + XY – 3Y 2 = 0 9x 2 – 6xy + y 2 + 18x – 6y + 8 = 0 represents two lines parallel to each other. Ans. Given equation is 10X2 + 6XY – 5XY – 3Y 2 = 0 2X(5X + 3Y) – Y (5X + 3Y) = 0 (2X – Y) (5X + 3Y) = 0 ∴ Either 2X – Y = 0 ⇒ 2(x + 1) – (y – 2) = 0 9x 2 – 6xy + y 2 +18x – 6y + 8 = 0 Comparing with ⇒ 2x – y + 4 OR, 5X + 3Y ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 a = 9, h = –3, b = 1, g = 9, f = –3, c = 8. Consider abc + 2 fgh – af 2 – bg 2 – ch 2 = 9 × 1 × 8 + ( –6 ) × 9 × –3 – 9 × 9 – 1 × 81 – 8 × 9 = 72 +162 – 81 – 81 – 72 = 0 = 0 ⇒ 5(x + 1)+ 3(y – 2) = 0 ⇒ 5x + 3y – 1 = 0 ∴ Th e g iven equation re presents two straight lines 2x – y + 4 = 0 and 5x + 3y – 1 = 0 Q -6) Find p and q if the equation =0 ∴ The given equation represents a pair of lines … (i) 12x2 + 7xy – py2 – 18x + qy + 6 = 0 represent a pair of perpendicular lines. Ans. 12x2 + 7xy – py2 – 18x + qy + 6 = 0 2 Now, h 2 – ab = ( –3 ) – 9 × 1 = 9 – 9 = 0 2 As h – ab = 0 lines are parallel. ∴ From (i) & (ii) 3 1 –1 ,g= ,f= ,c=k 2 2 2 a = 12, h = …(ii) 9x 2 – 6xy + y 2 +18x – 6y + 8 = 0 represents a pair of parallel straight lines. 7 q , b = –p, c = 6, g = –9, f = 2 2 Conditions of perpendicular lines, a+b 12 – p = 0 = 0 p = 12 Condition for pair of lines, abc + 2fgh – af 2 – bg 2 – ch2 = 0 Pair of Straight Lines Mahesh Tutorials Science 23 –240 + 2 – 2q 2 + 240 + 4q = 0 2 7 q q 12 × (–p)(6) + 2 (–9) × – 12 2 2 2 – (–p) 81 – 6 × 72 × – p – 108 – 49 =0 4 147 63q – 3q2 + 81p – =0 2 2 72 × – 12 – – 864 – 2q 2 – 4q + 2 = 0 147 63q – 3q2 + 81 × 12 – =0 2 2 147 63q – 3q2 + 972 – =0 2 2 q 2 – 2q + 1 = 0 (q – 1) (q – 1) = 0 i.e. q = 1 q 2 =1 ; q =1 ; p = 8, q = 1 Q -8) Find the value of k, if each of the following equations represents a pair of lines : i) 3x2 + 10xy +3y2 +16y +k = 0. ii) kxy + 10x +6y +4 = 0 iii) x2 + 3xy +2y2 +x – y +k = 0 Ans. i) Given equation is 147 63q – 3q 2 – =0 2 2 6q2 + 63q – 69 = 0 2q2 + 21q – 23 = 0 3x 2 + 10xy +3y2 +16y +k = 0. Comparing it with 2q2 + 23q – 2q – 23 = 0 q (2q + 23) –1 (2q + 23) = 0 ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 we get, a = 3, h = 5 b = 3 , g = 0, f = 8 c = k Since the equation represents a pair of lines –23 or q = 1 2 q= ∴ abc + 2 fgh – af 2 – bg 2 – ch2 = 0 ∴ (3)(3)(k) + 2 (8) (0) (5) –3(8)2 –3 (0)2 – k (5)2 p = 12 Q -7) Find p and q if the following equations represent a pair of parallel lines 2x 2 + 8xy + py2 + qx + 2y – 15 = 0 Ans. The equation is, comparing with Comparing it with ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 q , f = 1, c = –15 2 we get, a = 0, h = The lines are parallel Since the equation represents a pair of lines h – ab = 0 2 2 (4) p= k b = 0 , g = 5, f = 3 2 c=4 2 ( 4) ∴ 9k + 0 – 192 – 0 – 25k ∴ –16k – 192 = 0 ∴ –16k – 192 Hence k = –12 ii) Given equation is kxy + 10x + 6y + 4 = 0 2x 2 + 8xy + py 2 + qx + 2y – 15 = 0 a = 2, h = 4, b = p, g = =0 – 2( p) = 0 ∴ abc + 2 fgh – af 2 – bg 2 – ch2 = 0 = 2p k ∴ (0)(0)(4) + 2 (3) (5) –0 (3)2 – 0 (5)2 2 16 2 2 k – 4 = 0 2 p=8 The above equation represents pair of line abc + 2 fgh – af 2 – bg 2 – ch 2 = 0 2 q q 2 ( 8 ) ( –15 ) + 2 (1) ( 4 ) – 2 (1) – 8 2 2 2 – ( –15 ) ( 4 ) = 0 q 8 ( –30 ) + 2 (1) 4 – 2 – 2q 2 + 240 = 0 2 2 ∴ 0 + 15k – 0 – 0 – k 2 = 0 ∴ 15k –k2 = 0 ∴ –k (k – 15) = 0 ∴ k = 0 or k = 15 If k = 0 then the given equation becomes 10x + 6y + 4 = 0 which does not represent a pair of lines k≠ 0 Pair of Straight Lines Mahesh Tutorials Science 24 Hence k = 15 iii) Given equation is x 2 + 3xy +2y2 +x – y +k = 0 Comparing it with ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 we get, a = 1, h = 3 1 1 b=2,g= ,f=– 2 2 2 c=k Since the equation represents a pair of lines a h ∴ h b g f ∴ 3 2 1 2 1 – 2 ∴ – ac 2 bc 2 –0+0 – =0 4 4 ac 2 bc 2 – =0 4 4 ac2 + bc2 = 0 c2 (a + b) = 0 a + b = 0 or c = 0 Q-10) Find p and q if the equation px2 – 8xy + 3y2 + 14x + 2y + q = 0 represent two straight lines which are perpendicular. 1 2 1 – =0 2 2 ∴ 0– This is the required condition g f =0 c 3 2 1 c2 c bc ∴ a 0 – – 0 + 0 – =0 4 2 2 Also find the co-ordinate of their point of intersection. Ans. Given equation is px2 – 8xy + 3y2 + 14x +2y + q = 0 k Comparing it with ax2 + 2hxy + by2 +2gx + 2fy + c = 0, we get 2 3 1 1 ∴ 3 4 –1 = 0 2 1 –1 2k a = p, h = –4 , b = 3, g = 7 f = 1 and c = q Since the given equation represents a pair of line perpendicular to each other, 2 3 1 ∴ a+b=0 ∴ 3 4 –1 = 0 ∴ p+3=0 1 –1 2k ∴ p = –3 ∴ 2 (8k –1) –3 (6k +1 ) + 1 (–3 –4) = 0 ∴ 16k – 2 – 18k – 3 – 7 = 0 ∴ –2k –12 = 0 Also the equation represents a pair of straight lines ∴ –2k = 12 ∴ k = –6 a ∴ h g Q -9) Find the condition that equation ax2 +by +cx +cy = 0 may represent a l pair of lines Ans. Given equation is ax 2 +by +cx +cy = 0 with Ax 2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get A = a, H = 0, B = b, g = c c ,f= ,C=0 2 2 The given equation represents a pair of lines a 0 A H G H B G F F = 0 i.e, if 0 b C c c 2 2 Pair of Straight Lines c 2 c =0 2 0 h g b f =0 f c −3 −4 7 ∴ −4 3 1 = 0 7 1 q ∴ – 3 (3q – 1) + 4 (–4q – 7) + 7 (–4 – 21) = 0 ∴ –9q + 3 – 16q – 28 – 175 = 0 ∴ –25q – 200 = 0 ∴ – 25(q + 8) = 0 ∴ q = –8 ∴ p = –3 and q = –8 Mahesh Tutorials Science 25 BASIC ASSIGNMENTS (BA) : Q-2) Find the separate equations of the lines BA – 1 represented by the following equations : x2 – 4xy = 0 Q-1) Find the joint equation of the following pair of lines Ans. Given equation is x2 – 4xy = 0 ∴ x(x – 4y) = 0 i) 2x + y = 0 and 3x – 5y = 0 ii) Passing through (2, 3) and perpendicular ∴ The separate equations of lines are x = 0 and x – 4y = 0. to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 Q-3) Find the joint equation of pair of lines through Ans. i) Given equations : 2x + y = 0 and 3x – 5y = 0 Joint equation of the two lines u = 0 and v = 0 is uv = 0 the origin, which are perpendicular to the lines represented by 5x2 + 2xy – 3y2 = 0 Ans. Given equation is 5x2 + 2xy – 3y2 = 0. Comparing it with ax2 + 2hxy + by2 = 0, we ∴ Joint equation of 2x + y = 0 and 3x – 5y = 0 is get a = 5, h = 1, and b = – 3 Let m 1 and m 2 be the slopes of the lines (2x + y)(3x – 5y) = 0 ∴ 6x2 – 10xy + 3xy – 5y2 = 0 ∴ 6x2 – 7xy – 5y2 = 0 is the joint equation. represented by 5x2 + 2xy – 3y2 = 0 ∴ m1 + m2 = ii) Given equation of line is 3x + 2y – 1 = 0 m1.m2 = ∴ y=– 3 1 x+ 2 2 ∴ Slope is –3 2 –1 –1 Slopes of the required lines are m and m 1 2 x 2 + 3 3 ∴ Slope is Required lines also pass through the origin, therefore their equations are of the form –1 –1 y = m x and y = x m2 1 1 3 ∴ x + m1y = 0 and x + m2y = 0 ∴ The joint equation of the lines is given by Let u and v be the lines passing through (2, 3) and perpendicular to the above lines ∴ Slope of u is a –5 = b 3 Since, the required lines are perpendicular to these lines And equation of line is x – 3y + 2 = 0 ∴ 3y = x + 2 ∴ y= – 2h 2 = and b 3 2 and v is –3 3 ∴ Equation of u is y – 3 = (x + m1y)(x + m2y) = 0 ∴ x2 + (m1 + m2) xy + m1m2y2 = 0 ∴ x2 + 2 5 xy – y2 = 0 3 3 ∴ 3x2 + 2xy – 5y2 = 0 2 (x – 2) 3 ∴ 3y – 9 = 2x – 4 ....(i) ∴ 2x – 3y + 5 = 0 Equation of u is y – 3 = –3(x – 2) Q-4) Find k if, slope of one of the lines given by ∴ y – 3 = –3x + 6 ∴ 3x + y – 9 = 0 Ans. Given equation is kx2 + 4xy – y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0, we ....(ii) kx2 + 4xy – y2 = 0 exceeds the slope of the other by 8. ∴ Joint equation of (i) and (ii) is (2x – 3y + 5)(3x + y – 9) = 0 get a = k, 2h = 4, and b = – 1 Let m 1 and m 2 be the slopes of the lines ∴ 6x2 + 2xy – 18x – 9xy – 3y2 + 27y + 15x + 5y – 45 = 0 represented by kx2 + 4xy – y2 = 0. ∴ m1 + m2 = 4 and m1m2 = – k ∴ 6x2 – 7xy – 3y2 – 3x + 32y – 45 = 0 is the joint equation. According to the given condition, m2 = m1 + 8 ∴ m1+m1 + 8 = 4 Pair of Straight Lines Mahesh Tutorials Science 26 ∴ 2 m1 = – 4 ∴ m1 = – 2 Now, m1(m1 + 8) = – k ∴ ∴ b + 6h + 9a = 0 ∴ 9b + b + 6h = 0 This is the required condition. ∴ (–2)(–2 + 8) = – k ∴ –2(6) = – k ∴ –12 = – k ∴ k = 12 Q-2) If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is four times the other, Q-5) Find the joint equation of the lines passing through the origin having inclinations 600 and 1200 with X–axis Ans. Slope of the line having inclination 600 is m1 = tan 60 = 3 And Slope of the line having inclination 1200 0 is m2 = tan 1200 = tan (180 – 600) = – tan 600 =– 3 Required lines pass through the origin, therefore their equations are of the form y = mx ∴ Equations of the required lines are y = m1x and y = m2x i.e. y = show that 16h2 = 25ab Ans. Let m 1 and m 2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0. ∴ m1 + m2 = – According to the given condition, m2 = 4m1 ∴ m1 + 4m1 = – ∴ 5m1 = – 2h b ∴ m1 = – 2h 5b 3x+y=0 Also, m1(4m1) = ∴ The joint equation of these lines is given by ( 2h a and m1m2 = b b 3 x and y = – 3 x 3 x – y = 0 and i.e. b 2h + +a=0 9 3 3x – y )( ∴ 4m12 = ) 3x + y = 0 ∴ 3x2 – y2 = 0 ∴ m 12 = BA – 2 2h b ... (i) a b a b a 4b 2 Q-1) Find the condition that the line 3x + y = 0 may be perpendicular to one of the lines given by ax2 + 2hxy + by2 = 0 Ans. Auxiliary equation of ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0. Slope of the lines 3x + y = 0 is – 3 Now, one of the line s re pre se nte d by ax2 + 2hxy + by2 = 0 is perpendicular to the line 3x + y = 0 ∴ Slope of that line is m = ∴ m= 1 3 1 is a root of the auxiliary equation 3 bm2 + 2hm + a = 0. 2 a –2h ∴ = 5 b 4b ∴ a 4h 2 2 = 4b 25b ∴ 4h 2 a = , as b ≠ 0 25b 4 ∴ 16h2 = 25ab Q-3) Find the measure of acute angle between the lines represented by 2x2 + 7xy + 3y2 = 0 Ans. Given equation is 2x2 + 7xy + 3y2 = 0. Comparing with ax2 + 2hxy + by2 = 0, we get a = 2, h = 7 and b = 3 2 Let θ be the acute angle between the lines. ∴ tan θ 1 1 ∴ b + 2h + a = 0 3 3 = Pair of Straight Lines ... [From (ii)] 2 h 2 – ab = a +b 2 49 5 –6 2× 4 2 =1 = 2+3 5 Mahesh Tutorials Science 27 ∴ tan θ = 1 ∴ θ = 450 2 Q-4) Show that 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 represents a pair of lines, also find the 1 2 – 2 × ( – 3) 2 2 + ( – 3) = acute angle between them. Ans. Given equation of straight line is 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 Comparing with ax2 + 2hxy + by2 + 2gx + 2fy + 1 +6 4 –1 2 = c = 0, we get a = 2, h = – f= 1 , b = – 3, g = – 3, 2 = 19 , c = – 20 2 –2× ∴ tan θ = 5; = –2 25 4 5 2 ∴ θ = tan–1(5) If the equation represents a pair of lines then a h h b g f =0 g c f ∴ D = Q-5) Find the value of k, if the following equations represents a pair of lines 3x2 + 10xy + 3y2 + 16y + k = 0 a h g h b f g f c 1 2 –3 1 2 –3 19 2 –3 19 2 – 20 2 = Ans. Given equation is 3x2 + 10xy + 3y2 + 16y + k = 0 Comparing it with ax2 + 2hxy + by2 + 2gx + 2fy – – 361 –1 57 = 2 60 – – 10 + 4 2 2 –19 + ( – 3) – 9 4 + c = 0, we get a = 3, h = 5, g = 0, f = 8 and c = k Since the euation represents a pair of lines ∴ abc + 2fgh – af2 – bg2 – ch2 = 0 ∴ 3(3)(k) + 2(8)(0)(5) – 3(8)2 – 3(0)2 – k(5)2 = 0 ∴ 9k + 0 – 192 – 0 – 25k = 0 ∴ –16k – 192 = 0 ∴ –16k – 192 Hence, k = –12 BA – 3 = 2× = 240 – 361 1 20 + 57 + × 4 2 2 –19 – 36 – 3× 4 –121 77 165 + + 2 4 4 – 242 + 242 = 0 4 Since, determinant of the given equation is zero, it represents a pair of lines. = Q-1) If one of the lines given by 3x2 – kxy + ky2 = 0 is perpendicular to the line 5x + 3y = 0 Ans. Given equations is 3x2 – kxy + 5y2 = 0. Its auxiliary equation is 5m2 – km + 3 = 0 Slope of the line 5x + 3y = 0 is ∴ Slope of the perpendicular to 5x + 3y = 0 is Let θ be the acute angle between the pair lines. ∴ ∴ tan θ = 2 h 2 – ab a +b –5 3 3 . 5 3 is a root of auxiliary equation 5 5m2 – km + 3 = 0 2 3 3 ∴ 5 – k + 3 = 0 5 5 Pair of Straight Lines Mahesh Tutorials Science 28 Q-4) If the lines represented by ax2 + 2hxy + by2 45 3k – +3=0 ∴ 25 5 = 0 make angles of equal measure with the coordinate axes, then show that a = ± b. ∴ 45 – 15k + 75 = 0 ∴ 15k = 120 ∴ k=8 Ans. Let two lines make equal angle α with two coordinate axes. Q-2) Find the joint equation of pair of lines through the origin and perpendicular to the lines given by 2x2 – 3xy – 9y2 = 0 Ans. Given equation is 2x2 – 3xy – 9y2 = 0 i.e. 2x2 – 6xy + 3xy – 9y2 = 0 i.e. 2x(x – 3y) + 3y(x – 3y) = 0 i.e. (x – 3y)(2x + 3y) = 0 2 or α and π –α 2 π + α with the positive direction of 2 X-axis. Slope of the first line is m1 = tan α and 2 ∴ Seperate equations of 2x – 3xy – 9y = 0 are x – 3y = 0 and 2x + 3y = 0 1 –2 ∴ Slopes of these lines are and 3 3 Now, required lines are perpendicular to these lines, therefore their slopes are – 3 and Thus, these lines will form angle α and 3 . 2 π slope of the other line is m2 = tan – α or 2 π tan + α 2 ∴ m2 = cot α or m2 = –cot α ∴ m1m2 = tan α × cot α = 1 or m1m2 = tan α (–cot α) = –1 ∴ m1 m2 = ± 1 Since these lines also pass through the origin, their equations are of the form But m1m2 = 3 y = – 3x and y = x 2 ∴ a b a =±1 b ∴ a=±b i.e. 3x + y = 0 and 3x – 2y = 0 ∴ The combined equation of the lines is given by (3x + y)(3x – 2y) = 0 ∴ 9x2 – 3xy – 2y2 = 0 Q-3) Determine the nature of lines represented by x2 + 7xy + 2y2 = 0 Ans. Given equation is x2 + 7xy + 2y2 = 0 Comparing with ax2 + 2hxy + by2 = 0, we get a = 1, 2h = 7 i.e. h = 7 and b = 2 2 49 49 – 8 41 –2= = >0 4 4 4 ∴ The lines represented by x2 + 7xy + 2y2 = 0 are real and distinct. Now, h2 – ab = Q-5) Find the joint equation of pair of lines through (2, –3) and parallel to x2 + xy – y2 = 0 Ans. Solution is attached at end ADVANCED ASSIGNMENTS (AA) : AA – 1 Q-1) Find the joint equation of pair of lines throu gh the origin an d making an equilateral triangle with the line y = 3 Ans. Let OA and OB be the lines through the origin making an angle of 600 with the line y=3 ∴ OA and OB make an angle of 600 and 1200 with the positive direction of Xaxis. ∴ Slope of OA = tan 600 = 3 ∴ Equation of line OA is y = ∴ Pair of Straight Lines 3x–y=0 3x Mahesh Tutorials Science 29 ∴ (x – 1)(x – 6) = 0 ∴ The separate equations of x2 – 7x + 6 = 0 are Y B 60 0 60 y=3 A 0 and x–1=0 x–6=0 ... (i) ... (ii) Let the equation of side AB be x – 1 = 0 and equation of side CD be x – 6 = 0. 120 60 0 Consider, y2 – 14y + 40 = 0. ∴ (y – 4)(y – 10) = 0 0 600 X′ X O ∴ The separate equations of y2 – 14y + 40 = 0 are and Y′ Slope of OB = tan 1200 = tan (1800 – 600) = –tan 600 = – 3 ... (iii) ... (iv) Let the equation of side BC be y – 4 = 0 and equation of side AD be y – 10 = 0. Y ∴ Equation of line OB is y = – 3 x i.e. y–4=0 y – 10 = 0 A(1,10) D(6,10) B(1,4) C(6,4) y = 10 3x+y=0 ∴ The joint equation of the lines is given by ( 3x – y )( ) 3x + y = 0 i.e. 3x2 – y2 = 0 Q-2) Fin d the sepaeate equation s of the following lines 2 2 (x – 2) – 3(x – 2)(y + 1) + 2(y + 1) = 0 Ans. Given equation of line is O y=4 X x=1 x=6 (x – 2)2 – 3(x – 2)(y + 1) + 2(y + 1)2 = 0 ∴ (x – 2)2 – 2(x – 2)(y + 1) – (x – 2)(y + 1) Solving (i), (ii), (iii) and (iv), co-ordinates of vertices of the parallelogram are A (1,10), + 2(y + 1)2 = 0 ∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0 ∴ [(x – 2) – (y + 1)] [x – 2) – 2(y + 1)] = 0 B (1,4), C (6,4) and D (6,10). ∴ Equation of the diagonal AC is ∴ (x – 2 – y – 1)(x – 2 – 2y – 2) = 0 ∴ (x – y – 3)(x – 2y – 4) = 0 ∴ –5y + 50 = 6x – 6 ∴ 6x + 5y – 56 = 0 Thus, separate equation of the lines are x – y – 3 = 0 and x – 2y – 4 = 0 Q-3) Equation of pairs of opposite side of a parallelogram are x2 – 7x + 6 = 0 and y2 – 14y + 40 = 0. Find the joint equation of its diagonals. Ans. Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x 2 – 7x + 6 = 0 And the combined equation of sides BC and AD is y2 – 14y + 40 = 0. Consider x2 – 7x + 6 = 0 y – 10 10 – 4 6 = = x –1 1– 6 –5 Equation of diagonal BD is y – 4 4 – 10 –6 6 = = = 1– 6 –5 5 x –1 ∴ 5y – 20 = 6x – 6 ∴ 6x – 5y + 14 = 0 ∴ Equ atio ns o f th e di agon als of t he parallelogram are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0 respectively. ∴ Joint equation is (6x + 5y – 56)(6x – 5y + 14) = 0 ∴ 6x (6x – 5y + 14) + 5y (6x – 5y + 14) – 56 (6x – 5y + 14) = 0 Pair of Straight Lines Mahesh Tutorials Science 30 ∴ 36x2 – 30xy + 84x + 30xy – 25y2 + 70y – 336x + 280y – 784 = 0 ∴ 36x2 – 25y2 – 252x + 350y – 784 = 0 ∴ –3 m – 4 = 3 –3 1+ m – 4 ∴ 4m + 3 3 = 4 – 3m Q-4) Show that the joint equation of a pair of lines through the origin and each making an angle of α with x + y = 0 is x2 + 2sec 2axy + y2 = 0 Squaring both sides, we get Ans. Let OA and OB be the required lines. Let m be the slope of OA or OB. 2 ∴ Its equation is y = mx ... (i) it makes an angle α with the line x + y = 0 whose slope is –1. ∴ tan α = Since, required lines pass through the origin, their equations are of the form y = mx i.e. 2 tan α = ∴ 3(4 – 3m)2 = (4m + 3)2 ∴ 3(16 – 24m + 9m2) = 16m2 + 24m + 9 ∴ 48 – 72m + 27m2 = 16m2 + 24m + 9 ... (i) ∴ 11m2 – 96m + 39 = 0 m +1 1+ m ( –1) Squaring both sides, we get 2 3= (m +1) 2 (1 – m ) m= ∴ tan2 α (1 – 2m + m2) = m2 + 2m + 1 ∴ tan2 α – 2m tan2 α + m2 tan2 α = m2 + 2m + 1 ∴ (tan2 α – 1)m2 – 2(1 + tan2 α) m + (tan2 α – 1) = 0 2 y y ∴ 11 – 96 + 39 = 0 ... [From (i)] x x 11y 2 96y – + 39 = 0 ∴ x x2 e quati on w hich can be writte n as – 39x2 + 96xy – 11y2 = 0 1+ tan α ∴ m2 + 2 m + 1 = 0 2 1 – tan α 2 i.e. (9x2 + 48x2) + (24xy + 72xy) + (16y2 – 27y2) = 0 ∴ m2 + 2(sec 2α) m + 1 = 0 y y2 +1= ∴ 2 + 2(sec 2α) x x ... [From (i)] 2 ∴ y + 2xy sec 2α + x = 0 ∴ x2 + 2xy sec 2α + y2 = 0 is the required joint equation. (3x +4y)2– 3(4x – 3y)2 = 0 form an equilateral triangle. Ans. Slope of the line 3x + 4y = –5 is m1 = i.e. (9x2 + 24xy + 16y2) – 3(16x2 – 24xy + 9y2) = 0 i.e. (3x + 4y)2 – 3(4x – 3y)2 = 0 Hence, the line 3x + 4y + 5 = 0 and the lines AA – 2 –3 4 Let m be the slope of one of the line making an angle of 600 with the line 3x + 4y = –5. Since, the angle between the lines having slope m and m1 is 600, Pair of Straight Lines i.e. (9x2 + 24xy + 16y2) – (48x2 – 72xy + 27y2) = 0 (3x + 4y)2 – 3(4x – 3y)2 form an equilateral triangle. Q-5) Show that 3x + 4y + 5 = 0 and m – m1 tan 600 = 1+ m.m 1 y x ∴ 11y2 – 96xy + 39x2 = 0 ∴ 39x2 – 96xy + 11y2 = 0 is the required joint. 1+ tan2 α ∴ m2 – 2 m + 1 = 0 2 tan α – 1 2 ( 4m + 3 ) 2 ( 4 – 3m ) Q-1) If the slope of one of the lines given by ax2 + ab2 + 8h3 = 6abh is square of the slope of the other line, then show that a2b + ab2 + 8h3 = 6abh Ans. Let m be the slope of one of the lines given by ax2 + 2hxy + by3 = 0 ∴ Slope of the other line = m2 Mahesh Tutorials Science ∴ m + m2 = and m.m2 = 31 Comparing the coefficients of (i) and (ii), we –2h b ... (i) get a a i.e. m3 = ... (ii) b b a b 2b 2c = = = and 2a + c = 0 1 2 4 k ∴ a= ∴ (m + m2) = m3 + (m2)3 + 3(m)(m2)(m + m2) 3 a a 2 a –2h –2h = + 2 + ∴ b b b b b 3 ∴ ∴ a= ∴ 1= 2 – 8h a a 6ah = + 2 – 2 b b b3 b 2c and c = –2a k 2 ( – 2a ) k –4 k ∴ k=–4 3 Multiplying both sides by b , we get 3 2 2 –8h = ab + a b – 6abh ∴ a2b + ab2 + 8h3 = 6abh Q-2) If the lines given by ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + b)– 4h2 = 0 Ans. Lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1 ∴ Angle between the lines ax2 + 2hxy + by2 = 0 is 600 2 h 2 – ab tan 60 = a +b 0 2 3 = 2 h – ab a +b Q-4) Find the joint equation of bisectors of angles between the lines represented by 5x2 + 6xy – y2 = 0 Ans. Given equation is 5x2 + 6xy – y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0, we get a = 5, 2h = 6 and b = – 1 Let m1 and m2 be the slopes of the lines represented by 5x2 + 6xy – y2 = 0. ∴ m1 + m2 = 6 and m1.m2 = –5 ...(i) The separate equations of the lines are y = m1x and y = m2x, where m1 ≠ m2 ∴ m1x – y = 0 and m2x – y = 0 Let (x,y) be any point on one of the bisector of the angles between the lines. ∴ Distance of P from the line m1x – y = 0 is equal to the distance of P from line m2x – y = 0. ∴ 3(a + b)2 = 4(h2 – ab) ∴ 3(a2 + 2ab + b2) = 4h2 – 4ab Y Angle bisector ∴ 3a2 + 6ab + 3b2 + 4ab = 4h2 ∴ 3a2 + 10ab + 3b2 = 4h2 P(x,y) ∴ 3a2 + 9ab + ab + 3b2 = 4h2 ∴ 3a (a + 3b) + b(a + 3b) = 4h2 ∴ (3a + b)(a + 3b) – 4h2 = 0 × × Q-3) If the line x + 2 = 0 coincides with one of the lines represented by the Equation x2 + 2xy + 4y + k = 0 then prove that k = – 4 Ans. Given equation is X Angle bisector m1 x – y m2 x – y x2 + 2xy + 4y + k = 0 ... (i) One of the lines represented by given ∴ equation is x + 2 = 0. Let the other line be ax + by + c = 0 Squaring both sides, we get ∴ The combined equation of the lines is given by (x + 2) (ax + by + c) = 0 ∴ ax2 + bxy + cx + 2ax + 2by + 2c = 0 ∴ ax2 + bxy + (2a + c)x + 2by + 2c = 0 ...(ii) m12 +1 (m1 x – y ) m12 +1 = 2 = m 22 +1 (m 2 x – y ) 2 m22 +1 ∴ [(m22 + 1)(m1x – y)2] = [(m12 + 1)(m2x – y)2] Pair of Straight Lines Mahesh Tutorials Science 32 ∴ [(m22 + 1)(m12x2 – 2m1xy + y2) = (m12 + 1)(m22x2 – 2m2xy + y2) 2 2 2 ∴ m1 m2 x – 2m1m22xy + m22y2 + m12x2 – 2m1xy + y2 – m12m22x2 – 2m12m2xy + m12y2 + m22x2 – 2m2xy + y2 ∴ (m12 – m22)x2 + 2m1m2(m1 – m2)xy – 2(m1 – m2)xy – (m12 – m22)y2 = 0 Dividing throughout by ‘m1 – m2’, we get, (m1 + m2)x2 + 2m1m2xy – 2xy – (m1 – m2)y2 = 0 ∴ 6x2 – 10xy – 2xy – 6y2 = 0 ∴ 6x2 – 12xy – 6y2 = 0 ...[By (i)] ∴ x2 – 2xy – y2 = 0 is the required joint equation. Q-5) Find the measure of the acute angle between the lines represented by (a2 – 3b3) x2 + 8abxy + (b2 – 3a2)y2 = 0 Ans. Given equation is (a2 – 3b3) x2 + 8abxy + (b2 – 3a2)y2 = 0 Comparing with Ax2 + 2Hxy + By2 = 0, we get A = a2 – 3b2, H = 4ab and B = b2 – 3a2 Now, H2 – AB = 16a2b2 – (a2 – 3b2)(b2 – 3a2) = 16a2b2 – (a2 – 3b2)(3a2 – b2) = 16a2b2 – 3a4 – 10a2b2 + 3b4 = 3(a2 + b2)2 ∴ 3 (a2 + b2) H 2 – AB = Also, A + B = (a2 – 3b2)(b2 – 3a2) = –2(a2 + b2) Let θ be the acute angle between the lines. ∴ tan θ ( = ) 2 3 a2 + b2 2 H2 – AB = = 3 A + AB –2 a 2 + b 2 ∴ θ = 600 Pair of Straight Lines ( )