Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ p 1 2 2 2 2 bcos a b asin 0 ab bcos a b ab 2 2 2 2 2 2 2 2 b cos a sin b cos a sin Length of the perpendicular from point 2 2 2 2 2 2 bcos a b asin 0 ab bcos a b ab a b ,0 to line (2) is p2 2 2 2 2 2 2 2 2 b cos a sin b cos a sin On multiplying equations (2) and 3 , we obtain pp 2 2 2 2 bcos a b ab bcos a b ab 2 2 2 2 b cos a sin 2 2 2 2 bcos a b abbcos a b ab 1 2 2 2 2 2 2 b cos a sin ... 2 ...(3 2 2 2 bcos a b ab 2 2 2 2 2 b cos a sin 2 2 2 2 2 2 b cos a b a b 2 2 2 2 b cos a sin 2 2 2 4 2 2 2 a b cos b cos a b 2 2 2 2 b cos a sin 2 2 2 2 2 2 b a cos b cos a 2 2 2 2 b cos a sin 2 2 2 2 2 2 2 2 2 b a cos b cos a sin a cos 2 2 sin cos 1 2 2 2 2 b cos a sin 2 2 2 2 2 b b cos a sin 2 2 2 2 b cos a sin 2 2 2 2 2 b b cos a sin 2 2 2 2 b cos a sin 2 b Hence, proved. 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 ______________________________________________________________________________ Question 24: A person standing at the junction (crossing) of two straight paths represented by the equation 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time .Find equation of the path that he should follow. Solution 24: The equations of the given lines are 2x– 3y + 4 = 0 …(1) 3x + 4y – 5 = 0 …(2) 6x – 7y + 8 = 0 …(3) The person is standing at the junction of the paths represented by lines (1) and (2). 1 22 On solving equations (1) and (2), we obtain x and y 17 17 1 22 Thus, the person is standing at point , 17 17 The person can reach path (3) in the least time if he walks along the perpendicular line to (3) 1 22 from point , 17 17 . Slope of the line (3) = 6 7 1 7 Slope of the line perpendicular to line (3) = 6 6 7 1 22 The equation of the line passing through , 17 17 and having a slope of 7 is given by 6 22 7 1 y 17 x 6 17 6 17y 22 7 17x1 102 y132 **11**9 x7 **11**9x102 y125 Hence, the path that the person should follow is **11**9x102 y 125 . Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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