Question

What is the equation of a line that is perpendicular to 4x-3y=12 and passes through (-8,1)?

Answer 1

Answer:

y=-(3/4)x-5

Step-by-step explanation:

First we must write our equation in slope intercept form:

$y=mx+b$

Where m is slope and b is the y-intercept.

$4x-3y=12\\\\-3y=-4x+12\\\\y=\frac{4}{3}x-4$

Now we will find a line that is perpendicular to $y=\frac{4}{3}x-4$

To do this we will get the negative inverse of our original slope which is 4/3 and the negative inverse is: $-\frac{3}{4}$.

Now we must determine our y-intercept for the perpendicular line by using the slope calculated above and plugging it in to the following equation:

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

Where m is the slope, x1 = -8 and y1=1 and so:

$y=-\frac{3}{4}(x-(-8))+1\\\\y=-\frac{3}{4}x-\frac{24}{4}+1\\\\y=-\frac{3}{4}x-6+1\\\\y=-\frac{3}{4}x-5$