Write an equation of the line that passes through the given point and is perpendicular to the graph of the given equation?

1. (-9, 3); 3x + y = 5

2. (-8, 3); y +4 = -2/3(x-2)

3. (3, -6); x + y = -4

Let's take a look at the frist problem. 2 and 3 can be solved the same way.

First of all you have to rewrite your equation 3x + y = 5 (this is a standard form) in slope-intercept form by subtracting 3x from both sides of the equation. We have y = -3x +5. That means the sloppe of the line is

m=-3 (if we use y = mx+b form of the linear equation). The line which is perpendicular to that will have the slope equal to the reciprocal of (-3) taken with negative sign. Thus, the new slope equals 1/3. Or your equation for the perpendicular line should be like

y = (1/3)x +b

In order to find "b" use the point given in the problem. If x =-9 y =3. Thus, we have

3 = (1/3)(-9) +b = -3 +b or b = 3+3 =6

The equation for a perpendicular line becomes

y = (1/3)x +6

Repeat the same steps for problems 2 and 3.

## Comments

Thank you both!