Question

Find the slope-intercept form for the line perpendicular to 2x - 3y = -6 and passing through (4, 9)

Answer 1

The vector $(2,-3)$ is a normal vector of the line $2x-3y=-6,$ and hence it is a collinear vector of the required line. Therefore, we get the equation of the line in the form $\frac{x-4}{2}=\frac{y-9}{-3},$ which is equivalent to

$-3x+12=2y-18.$

We conlcude that the slope-intercept form for the line perpendicular to 2x - 3y = -6 and passing through (4, 9) is

$y=-\frac{3}2x+15.$