Answer:
[tex]y = \frac{3}{4}x - 8[/tex]
Step-by-step explanation:
We know that the slope of a line perpendicular to a line with a given slope is the negative reciprocal of that given slope. We are given that we want a line perpendicular to that of 4x + 3y = 3. To find the slope of this given line, we put it into slope intercept form y = mx + b where m is the slope:
[tex]4x + 3y = 3[/tex]
[tex]3y = 3 - 4x[/tex]
[tex]y = 1 - \frac{4}{3}x[/tex]
So we have that the slope of our given line is -4/3. The negative reciprocal of this is 3/4. We thus have that the slope of the line we are looking for is 3/4.
We then know that the equation of the line we want fits the form [tex]y=\frac{3}{4} x + b[/tex], which we know through point slope form. We are then given that (8, -2) is a point on this line. Plugging in:
[tex]-2 = \frac{3}{4}(8) + b[/tex]
[tex]-2 = 6 + b[/tex]
[tex]b = -8[/tex]
We then get the equation of the line we are looking for as [tex]y = \frac{3}{4}x - 8[/tex].
