Question

Given that 3y = x + 6 and y = kx + 12 are perpendicular, find the value of k

Answer 1

k = -3

Step-by-step explanation:

Let's rewrite the equations in their slope-intercept forms:

$y = \frac{1}{3}x + 2\:\:\:\:\:\:\:(1)$

$y = kx + 12\:\:\:\:\:(2)$

In order for Eqn(2) to be perpendicular to Eqn(1), the slope of Eqn(2) must be the negative reciprocal of that of Eqn(1). Therefore, $k = -3$.