Question

What is the equation of the line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5)?
y=x+9
y=-3x-4
y=-x-13
y-x-1

Answer 1

Rewriting the equation of the given line in slope-intercept form,

$2x-3y=13\\\\2x-13=3y\\\\y=\frac{2}{3}x-\frac{13}{3}$

Thus, the slope of the given line is 2/3. Since perpendicular lines have slopes that are negative reciprocals of each other, we want to find the line with a slope of -3/2.

Substituting into point-slope form,

$y-5=-\frac{3}{2}(x+6)\\\\y-5=-\frac{3}{2}x-9\\\\\boxed{y=-\frac{3}{2}x-4}$