Help with Parallel and Perpendicular Lines

Write the equation of the line that is perpendicular to the line y = 3/2 x + 4 and passes through the point (-6, 3)

A. y = 3/2 x - 1
y = 3/2 x + 3
y = - 3/2x - 1
y = - 3/2x + 3




By the way i am online schooled, So i don't have a teacher explain the stuff to me, i have to read on the website, which i can't find it. :/
Help please? and explain how to do it..? cause i have 3 other questions like this.

If it is perpindicular, that means that the slope is the negative reciprocal of the original slope.
In this case:
original slope = 3/2
perpindicular slope = -2/3

Now it also passes though (-6, 3)

y = -2/3x + b
3 = (-2/3)*-6 + b
3 = 4 + b
-1 = b

So the line should have the form:
y = -2/3x - 1

Which looks like it should be answer C, im guessing that you wrote it down something down wrong.

If you are given a slope :-
1. to find perpendicular slope = its always negative reciprocal slope

In this case:
original = 3/2
perpendicular = -2/3

equation of a line ---> y = mx + b

substitute the point (-6,3) for x & y and the perpendicular slope for m
this results in :-

3 = (-2/3)*(-6) + b ---------> solve the equation for b (the y-intercept)
3 = 4 + b
-1 = b

y = -2/3x - 1