Question

What is an equation of the line that passes through the point (8, -8) and isperpendicular to the line 4x - 3y = 18?

Answer 1

Solution:

Let us denote by L1 the line given by the following equation:

$4x-3y=18$

solving for 3y, this equation is equivalent to:

$4x-18=3y$

that is:

$3y=4x-18$

now, solving for y, we obtain:

$y=\frac{4}{3}x-\frac{18}{3}=\frac{4}{3}x-6$

that is:

$y=\frac{4}{3}x-6$

now, perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is

$m=-\frac{3}{4}$

with this information, we can say that the provisional slope-intercept form of the perpendicular line to L1 is:

$y\text{ =-}\frac{3}{4}x+b$

our goal is now to find the y-intercept b of this line. To do this, we can replace the point (x,y)=(8, -8) into the previous equation, to get:

$-8\text{ =-}\frac{3}{4}(8)+b$

solving for b, we get:

$b=-8+\frac{3}{4}(8)=-2$

so that, we can conclude that the equation of the line would be:

$y\text{ =-}\frac{3}{4}x-2$