Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ Question 8: Find the area of the triangle formed by the line y – x = 0, x + y = 0 and x – k = 0. Solution 8: The equation of the given lines are y – x = 0 …(1) x + y = 0 …(2) x – k = 0 …(3) The point of intersection of lines (1) and (2) is given by x = 0 and y = 0 The point of intersection of lines (2) and (3) is given by x = k and y = -k The point of intersection of lines (3) and (1) is given by x =k and y = k Thus, the vertices of the triangle formed by the three given lines are (0, 0) ,(k, -k), and (k, k). , x , y , and x , y is We know that the area of a triangle whose vertices are x y , 1 1 1 1 2 3 2 3 1 3 1 2 2 x y y x y y x y y Therefore, area of the triangle formed by the three given lines 1 0 k k k k 0 k 0 k square units 2 1 2 2 k k square units 2 1 2 2 k square units 2 2 k square units 2 2 3 3 Question 9: Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point. Solution 9: The equation of the given lines are 3x + y – 2 = 0 …(1) px + 2y – 3 = 0 …(2) 2x – y – 3 = 0 …(3) On solving equations (1) and (3), we obtain X = 1 and y = -1 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2). P (1) + 2(-1) -3 = 0 P – 2 – 3 =0 P = 5 Thus, the required value of p is 5. Question 10: If three lines whose equations are y m1 x c1, y m2 x c2, and y m2x c2 are concurrent, then show that m c c m c c m c c 1 2 3 2 3 1 3 1 2 0 Solution 10: The equation of the given lines are y m x c …(1) 1 1 y m x c …(2) 2 2 y m x c …(3) 2 3 On subtracting equation (1) from (2) we obtain 0 m m x c c 2 1 2 1 m m x c c 1 2 2 1 c2 c1 x m m 1 2 On substituting this value of x in (1), we obtain c2 c1 y m1 c1 m1 m2 m1 c2 m1 c1 y c1 m1 m2 m1 c2 m1 c1 m1 c1 m2c1 y m1 m2 m1 c2 m2c1 y m m 1 2 c c m c m c , is the point of intersection of line (1) and (2). 2 1 1 2 2 1 m1 m2 m1 m2 It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3). Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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