# Equations of Perpendicular Lines

Time to surf the *x* and *y* vibes and find out what the coordinate plane has in store for perpendicular lines. Sure, the *x*- and *y*-axes are perpendicular and those graph paper squares are totally perpendicular, but how do we find the equations of lines that are perpendicular to other lines? And we aren't talking your typical *y* = 3. We mean weird lines, with slopes of ¼ and *y*-intercepts like -23.

The secret to both parallel and perpendicular lines on the coordinate plane is the slope. Parallel lines have the same slope and they never intersect. Since perpendicular is sort of the opposite of parallel, we'd expect the slopes to be opposite some way. What we might not expect is that the slope is opposite in *two* ways.

For two lines to be perpendicular, one line has to have a slope that is the negative reciprocal of the slope of the other.

Say what?

Okay, think of it this way. We've got a line *y* = 2*x* + 1.

We want a line that intersects this line at a right angle. How do we get that? For one thing, the line has to slope in the opposite direction, so let's try going from a 2 to -2.

They definitely intersect, but it still doesn't look quite right, does it? The angles formed aren't 90° angles…yet. Basically, to form a right angle, we need to get nitty gritty. A line with a slope of 2 goes up 2 units on the *y*-axis for every 1 on the *x*-axis.

A perpendicular line should go down 1 unit for every 2 units on the *x*-axis. This means a slope of not just -2, but -½. That's the negative reciprocal.

So let's take a really weird line like *y* = ¼*x* – 23. A perpendicular line would be *y* = -4*x* – 23. Or *y* = -4*x* + 1, or *y* = -4*x* + 1001. The *y*-intercept doesn't really matter because it only changes *where* the two lines intersect, not *how*.

### Sample Problem

Find a line perpendicular to the line 2*x* + *y* = 7 that goes through the point (4, 5).

We've gotta find the slope of the line, so let's change this sucker into slope-intercept form. It'll look something like *y* = -2*x* + 7. Actually, it'll look exactly like that.

After we pluck the slope -2 out from the equation, we need its negative reciprocal. The reciprocal means flipping the number so its numerator and denominator are switched, and then taking its negative. In this case, we end up with -½. (Two wrongs might not make a right, but in this case, two negatives certainly help that process along.)

To get the new equation, we start with *y* = ½*x* + *b*. To find what *b* equals, we'll need to plug in that point (4, 5).

*y* = ½*x* + *b*

5 = ½(4) + *b*

5 = 2 + *b**b* = 3

So, our perpendicular line is *y* = ½*x* + 3 or 2*y* – *x* = 6.

### Sample Problem

Are *x* + *y* = 5 and *x* – *y* = 5 perpendicular?

In other words, this question asks whether or not the slopes of these two lines are negative reciprocals of each other. That's really all it takes for two lines to be perpendicular, right?

In slope-intercept form, *x* + *y* = 5 turns into *y* = -*x* + 5, which has a slope of -1. The other equation, *x* – *y* = 5, turns into *y* = *x* – 5, which has a slope of 1.

So, is 1 the negative reciprocal of -1? You betcha. They're as perpendicular as can be.