In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

line 2x − 3y = 9, point (−4, 0)

Equation of the line is: 2x - 3y = 9 and point is (-4, 0).

We know that the slope of two perpendicular line is negative reciprocal of each other.

It means if slope of the original line is mthen the slope of the perpendicular line is -1 / m.

We will use slope - point formula:
$y-y_1=m(x-x_1)$

We have given equation of the line is,

2x - 3y = 9

Rewriting it in to the slope - intercept form,

$y=\frac{2}{3}x-3$

Comparing it with y = mx + b we get slope of the original line is $\frac{2}{3}$.

We know that the slope of the perpendicular line is negative reciprocal of the original line.

So slope of the perpendicular line is $-\frac{3}{2}$.

Using slope - point formula,

$y-y_1=m(x-x_1)$

Substituting the given point and slope,

role="math" localid="1644460257179" $y-0=-\frac{3}{2}(x+4)$

Simplifying it,

$y=-\frac{3}{2}x-6$.

Hence, equation of the perpendicular line to 2x - 3y = 9 and passing though the point (-4, 0) is $y=-\frac{3}{2}x-6$.