Question

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

Answer 1

y=-(3/4)x-5

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=(4)/(3)x-4`

Now we will find a line that is perpendicular to
`y=(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:
`-(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=-(3)/(4)(x-(-8))+1\\\\y=-(3)/(4)x-(24)/(4)+1\\\\y=-(3)/(4)x-6+1\\\\y=-(3)/(4)x-5`