Find the equation of the line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and perpendicular to the line 4x – 2y = 3.
Straight Line Eq
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Question

Find the equation of the line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and perpendicular to the line 4x – 2y = 3.

KnowledgeBoat · Accepted Answer

Given line equations are, ⇒ 7x + 6y - 71 = 0 ⇒ 7x + 6y = 71 .......(1) and ⇒ 5x - 8y + 23 = 0 ⇒ 5x - 8y = -23 .......(2) Multiplying (1) by 4, we get : ⇒ 28x + 24y = 284 .......(3) Multiplying (2) by 3, we get : ⇒ 15x – 24y = -69 .......(4) On adding (3) and (4), we get : ⇒ 28x + 24y + 15x - 24y = 284 + (-69) ⇒ 28x + 15x + 24y - 24y = 284 - 69 ⇒ 43x = 215 ⇒ x = ⇒ x = 5. Substituting value of x in equation (2), we get : ⇒ 5.5 - 8y = -23 ⇒ 25 - 8y = -23 ⇒ 8y = 25 + 23 ⇒ 8y = 48 ⇒ y = 6. Hence, the required line passes through the point (5, 6). Given, ⇒ 4x – 2y = 3 ⇒ 2y = 4x – 3 ⇒ y = 2x – . Comparing above equation with y = mx + c we get, Slope (m) = 2 Let slope of required line be m1. As, the lines are perpendicular to each other so product of their slopes = -1. ⇒ m × m1 = -1 ⇒ 2 × m1 = -1 ⇒ m1 = . Thus, equation of the line with slope and passing through (5, 6) is : ⇒ y – y1 = m(x – x1) ⇒ y – 6 = (x – 5) ⇒ 2(y – 6) = -1(x - 5) ⇒ 2y - 12 = -x + 5 ⇒ 2y + x = 5 + 12 ⇒ x + 2y = 17. Hence, equation of required line is x + 2y = 17.

Mathematics

Find the equation of the line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and perpendicular to the line 4x – 2y = 3.

Straight Line Eq

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