Find k, if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = 4?
Slope of line 2x - 3y = 11 is $$\frac{-2}{-3}$$
= $$\frac{2}{3}$$
Slope of line 3x + ky = 4 is $$\frac{-3}{k}$$
Also, product of slopes of two perpendicular lines is -1
=> $$\frac{2}{3} \times \frac{-3}{k} = -1$$
=> $$\frac{-2}{k} = -1$$
=> $$k = -2 \times -1 = 2$$
=> Ans - (D)
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