Q.1. Find the value of k so that the straight line 2x + 3y + 4 + k(6x - y + 12) = 0 is perpendicular to the line 7x + 5y - 4 = 0. Share with your friends Share 1 Varun Rawat answered this The given equation of the first line is, 2x + 3y + 4 + k6x - y + 12 = 0⇒2 + 6kx + 3-ky + 4 + 12k = 0⇒ 3-ky = -2+6kx - 4+12k = 0⇒y = 2+6kxk-3 +12k+4k-3 .....1Slope of line 1 is m1 = 2 + 6kk - 3The equation of the other line 2 is, 7x + 5y - 4 = 0⇒5y = 4 - 7x⇒y = -7x5 + 45 .....2Slope of line 2 is m2 = -75Since, line 1 is ⊥ to line 2, then m1 m2 = -1⇒2 + 6kk - 3 × -75 = -1⇒14 + 42k5k - 15 = 1⇒42k + 14 = 5k - 15⇒37k = -29⇒k = -2937 0 View Full Answer
