Hi Kristy. I can show you how it's done with a similar problem, then you can follow those steps in solving your question.

Example: Find the equation of the line that is perpendicular to 2x = y - 5 and that passes through the point (4, -3). Write the equation in standard form.

Solution:
Step 1: Find the slope of the equation given. I want a line that is perpendicular to it, so I need to find the "negative reciprocal" of the other line's slope. To find the slope of 2x = y - 5, I first put it in "slope-intercept" form (that is, y = mx + b form):

2x = y - 5
y - 5 = 2x
y = 2x + 5

Now I see that the slope of that line is 2 (the number 2 takes the place of m in y = mx + b). So the slope of the new line that is perpendicular to it is the negative reciprocal of 2. So I change the sign: -2. Then I take the reciprocal of that: (-1/2). Now I know the slope of the new line.

To write the equation of a line, all you need is one point it passes through and the slope. That's because the point is (x, y) and the slope is m, so when you substitute these values into y = mx + b, you just have to solve for b. So for my line, I know m = (-1/2) and x = 4 and y = -3.

y = mx + b
-3 = (-1/2)(4) + b
-3 = -2 + b
b = -1

Therefore the equation of the new line is y = (-1/2)x - 1. I have substituted in the slope m and the b value I just found.

The last step is to write this in "standard form" That means the form "Ax + By + C = 0" where A, B, and C are all whole numbers and A is positive. So really, I just move everything to the left side and multiply by 2 to get rid of the fraction:

y = (-1/2)x - 1

(1/2)x + y + 1 = 0

1x + 2y + 2 = 0. <--- standard form

That's the solution for my example. Use the same steps to solve your question, Kristy.

Hope this helps,
Stephen La Rocque.