Answer:
[tex]y=\frac{3}{8}x+11[/tex]
Step-by-step explanation:
To find a line that is perpendicular to 8x + 3y = -6 and goes through (-24, 2), lets first find what the line's slope would be.
We can find this by finding the slope of 8x + 3y = -6 and taking the negative reciprocal of it.
We can find the slope of that line by putting it in slope-intercept form:
8x + 3y = -6
Subtract 8x from both sides.
3y = -6 - 8x
Divide both sides by 3.
[tex]y=-\frac{6}{3}-\frac{8}{3}x[/tex]
[tex]y=-\frac{8}{3}x-2[/tex]
So the slope of that line would be -8/3.
The negative reciprocal of -8/3 would be 3/8.
Now we know that the new line would have to pass through the point (-24, 2). We can use this point and write the equation in point-slope form:
[tex]y-2=\frac{3}{8}(x+24)[/tex]
Now lets change this into slope-intercept form. Add 2 to both sides.
[tex]y=\frac{3}{8}(x+24) +2[/tex]
Distribute the 3/8.
[tex]y=\frac{3}{8}x+9+2[/tex]
Simplify.
[tex]y=\frac{3}{8}x+11[/tex]
And now we have our equation in slope-intercept form.
I hope you find this helpful.
