Lesson Video: Equations of Parallel and Perpendicular Lines Mathematics • 11th Grade

Video Transcript

In this video, we will learn how to write the equation of a line parallel or perpendicular to another line. We’ll consider cases where we already know the slope. And we’ll also look at cases where we’re given two points and we need to find the slope of that line before finding the slope of a parallel or a perpendicular line.

But first, let’s review a few things about lines. 𝑦 equals 𝑚𝑥 plus 𝑏 is a general form for a straight-line equation. And in particular, we call this the slope–intercept form. In this form, the coefficient of 𝑥 — the 𝑚 variable — is the slope. And 𝑏 equals the 𝑦-intercept, the place where this function crosses the 𝑦-axis. As we said on the first slide, we want to consider parallel lines and perpendicular lines.

We know parallel lines never intersect one another; they never cross. We usually represent that with a small triangle on either line. And occasionally, when there’s more than one pair of parallel lines, you might see them noted with a multiple of the little triangles. In that case, the parallel lines will be the pair with the matching symbols. What we’re showing are two sets of parallel lines.

It’s also worth noting that you might see this parallel symbol used to represent parallel lines. Here, line segment 𝐴𝐵 is parallel to line segment 𝐶𝐷. But the most important thing to remember about parallel lines is that they have the same slope. If we look at straight-line equations given in slope–intercept form and they have the same value for the variable 𝑚 for the coefficient of 𝑥, they will be parallel.

Now, let’s consider what perpendicular lines are. Perpendicular lines intersect at a 90-degree angle. And this is often noted with the right angle symbol. Now, the set of perpendicular lines that I’ve drawn here are horizontal and vertical. But this is not always the case. Perpendicular lines can occur at any orientation. And when it comes to the slope of perpendicular lines, they are negative reciprocals of one another. So, parallel lines have the same slope; their 𝑚 values are the same. And in perpendicular lines, if one of the lines has a slope of 𝑚, the other line will have the negative reciprocal, which is negative one over 𝑚. It’s also worth noting here that there are lines that intersect that are neither parallel or perpendicular. Lines that intersect forming any angle besides a right angle do not fit in the category parallel or perpendicular.

Let’s use this information to start working with the equations of parallel and perpendicular lines.

Determine whether the lines 𝑦 equals negative one-seventh 𝑥 minus five and 𝑦 equals negative one-seventh 𝑥 minus one are parallel, perpendicular, or neither.

The categories parallel, perpendicular, or neither are always, we categorize, intersections of lines. Parallel lines do not intersect. Perpendicular lines intersect at a 90-degree angle. The category neither here represents all the lines that do intersect but do not form a 90-degree angle. Parallel, perpendicular, or neither.

But we’re not given a graph for these two lines. Of course, we could try and draw a graph for both of these lines. But we can determine parallel, perpendicular, or neither without graphing these two equations. Both of these straight lines are given in the form 𝑦 equals 𝑚𝑥 plus 𝑏. In both cases, the coefficient of 𝑥 — the 𝑚 variable — is negative one-seventh. The 𝑚 variable represents the slope. And so, we can say that the slope of line one is negative one-seventh and the slope of line two is negative one seventh, which reminds us, “parallel lines have the same slope.” That is why they do not intersect. Since both of these lines have a slope of negative one-seventh, we can classify them as parallel lines without graphing.

In this example, we’ll again be classifying lines. But this time we’re not given the equations of the lines. We’re only given two points that lie on each of the lines.

Given that the coordinates of the points 𝐴, 𝐵, 𝐶, and 𝐷 are negative 15, eight; negative six, 10; negative eight, negative seven; and negative six, negative 16, respectively, determine whether line 𝐴𝐵 and line 𝐶𝐷 are parallel, perpendicular, or neither.

Points 𝐴 and 𝐵 fall on the line 𝐴𝐵 and points 𝐶 and 𝐷 fall on the line 𝐶𝐷. To classify the lines, we have to remember: parallel lines have the same slope and they do not intersect. Perpendicular lines have negative reciprocal slopes and intersect at a 90-degree angle. And neither are lines that are not parallel or perpendicular, lines that do intersect but do not form a right angle. This means to consider whether or not these lines are parallel or perpendicular, we need to know the slopes of these lines.

In the general form, 𝑦 equals 𝑚𝑥 plus 𝑏, the 𝑚 represents the slope. And we can find the slope 𝑚 if we have two points by saying 𝑚 equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. In order to categorize these lines, we need to find the slopes of line 𝐴𝐵 and line 𝐶𝐷. We can start with line 𝐴𝐵. Let point 𝐴 be 𝑥 one, 𝑦 one and point 𝐵 be 𝑥 two, 𝑦 two. Then, the slope will be 10 minus eight over negative six minus negative 15. 10 minus eight is two. Negative six minus negative 15 is negative six plus 15, which is positive nine. So, we can say that the slope of line 𝐴𝐵 is two-ninths.

We repeat this process for line 𝐶𝐷. Let 𝐶 be 𝑥 one, 𝑦 one and 𝐷 be 𝑥 two, 𝑦 two. And we’ll get 𝑚 equals negative 16 minus negative seven over negative six minus negative eight. Negative 16 minus negative seven is negative 16 plus seven, which is negative nine. Negative six minus negative eight is negative six plus eight which is two. The slope of line 𝐶𝐷 is then negative nine over two.

If we compared these two slopes, negative nine over two is the negative reciprocal of two over nine. And if you weren’t sure, you can multiply them together. Reciprocals multiply together to equal one and negative reciprocals multiply together to equal negative one. These two slopes are the negative reciprocals of one another, making these lines perpendicular.

Here’s another example.

Which axis is the straight line 𝑦 equals three parallel to?

First, we know that parallel lines do not intersect and they have the same slope. We have the equation 𝑦 equals three. If we think about the general form for a straight line 𝑦 equals 𝑚𝑥 plus 𝑏 and we have 𝑦 equals three, which means that we have a slope of zero. From there, if we sketch a 𝑦- and 𝑥-axis, graphing the line 𝑦 equals three would look like this. We should notice here that the line 𝑦 equals three crosses the 𝑦-axis. And we know that’s true because it has a 𝑏-value of three. And in the general form, the 𝑏 represents the 𝑦-intercept.

We can, therefore, say that the straight line 𝑦 equals three is not parallel to the 𝑦-axis because it intersects the 𝑦-axis. But we can say that 𝑦 equals three is parallel to the 𝑥-axis. The 𝑥-axis is the line where 𝑦 equals zero. The straight line 𝑦 equals three is parallel to the 𝑥-axis.

In the next example, we need to find the equation of a line if we’re given a point on that line and then two points on a line perpendicular to that line.

Determine, in slope–intercept form, the equation of the line passing through 𝐴: 13, negative seven perpendicular to the line passing through 𝐵: eight, negative nine and 𝐶: negative eight, 10.

So, here’s what we’re thinking. We have points 𝐵 and 𝐶, which form a line. Point 𝐴 does not fall on this line. But point 𝐴 falls on a line perpendicular to the line 𝐵𝐶. And we’re trying to find in slope–intercept form the equation of the line that passes through point 𝐴. Slope–intercept form is the form 𝑦 equals 𝑚𝑥 plus 𝑏. That means we need the slope of this line and the 𝑦-intercept. But since we don’t know two points along this line, we’ll have to find the slope a different way.

We remember perpendicular lines have negative reciprocal slopes. And since we do know two points along the line 𝐵𝐶, we can find the slope of line 𝐵𝐶. And the slope along the line that includes point 𝐴 will be equal to negative one over the slope of the line from 𝐵𝐶. This is just a mathematical way to say that these two values will be negative reciprocals of one another. This means our first job is to find the slope of line 𝐵𝐶. If we know two points along the line, we can find their slope by taking 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one.

We’ll let 𝐵 be 𝑥 one, 𝑦 one and 𝐶 be 𝑥 two, 𝑦 two. And the slope of line 𝐵𝐶 will be equal to 10 minus negative nine over negative eight minus eight. 10 minus negative nine is 19. And negative eight minus eight equals negative 16. We can say that the slope of line 𝐵𝐶 is 19 over negative 16. But more commonly, we would include the negative sign in the numerator and say the slope of line 𝐵𝐶 equals negative 19 over 16. The slope of the line containing point 𝐴 is the negative reciprocal of this value.

To find the reciprocal of a fraction, we flip it. The reciprocal of negative 19 over 16 is 16 over negative 19. But we have to be careful here because we need the negative reciprocal. And that means negative 16 over negative 19 simplifies to 16 over 19. The slope of the line passing through point 𝐴 is then 16 over 19. At this point, we have the slope of the line passing through point 𝐴. And we have one point that falls along that line.

To find the 𝑦-intercept form of this equation, we could then use the point–slope formula, which says 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one, where 𝑥 one, 𝑦 one is a point along the line. Point 𝐴 is 𝑥 one, 𝑦 one. And so, we have 𝑦 minus negative seven equals 16 over 19 times 𝑥 minus 13. Minus negative seven is plus seven. We distribute that 16 over 19 times 𝑥. And 16 over 19 times negative 13 equals negative 208 over 19.

Since we want the equation in slope–intercept form, we need to get 𝑦 by itself by subtracting seven from both sides. Negative 208 over 19 minus seven is negative 341 over 19. This line 𝑦 equals 16 over 19𝑥 minus 341 over 19 is perpendicular to the line 𝐵𝐶 and passes through point 𝐴.

Here’s another example involving parallel lines.

The straight lines eight 𝑥 plus five 𝑦 equals eight and eight 𝑥 plus 𝑎𝑦 equals negative eight are parallel. What is the value of 𝑎?

We know that parallel lines have the same slope. And in slope–intercept form, 𝑦 equals 𝑚𝑥 plus 𝑏, the coefficient of the 𝑥-variable 𝑚 represents the slope. We’ve been told that these two lines are parallel. And that means they will have the same slope. To find the slope of these lines, we’ll convert them to slope–intercept form. To do that, we get 𝑦 by itself. Since both equations have eight 𝑥 on the left, we’ll subtract eight 𝑥 from both sides of both equations. On the left, we would have five 𝑦 equals negative eight 𝑥 plus eight. And on the right, 𝑎𝑦 equals negative eight 𝑥 minus eight. We need to get 𝑦 by itself for slope–intercept form. So, we can divide through by five. And the equation on the left in slope–intercept form will be 𝑦 equals negative eight-fifths 𝑥 plus eight-fifths.

On the right, we need to do something similar. To get 𝑦 by itself, we’ll divide through by 𝑎. And our second equation will be 𝑦 equals negative eight over 𝑎𝑥 minus eight over 𝑎. These slopes have to be equal to each other if these lines are parallel. Since one of the slopes is negative eight over five, the other slope will need to be negative eight over five. And that tells us that 𝑎 must be five for these two lines to be parallel. If we go back in and plug in five for 𝑎, we see that the ratio of coefficients between these two equations are equal to each other, which makes them parallel.

In our final example, we’re given three points that form a right-angled triangle. And we’ll use what we know about parallel or perpendicular lines to solve for a missing value in one of the points.

Suppose that the points 𝐴: negative three, negative one; 𝐵: one, two; and 𝐶: seven, 𝑦 form a right-angled triangle at 𝐵. What is the value of 𝑦?

We can go ahead and make a sketch of these points. 𝐴 is negative three, negative one. 𝐵 is one, two. We know the 𝑥-coordinate of point 𝐶 is seven. And that means 𝐶 will be located somewhere along this line. We know the line 𝐴𝐵. And we’ve been told that the right angle of this triangle is at point 𝐵. We can get a general idea of where we think point 𝐶 would be. But this is not a good way to find an accurate answer. But because we know this is a right triangle, we could say that line 𝐴𝐵 is perpendicular to line 𝐵𝐶. That means the slope of line segment 𝐵𝐶 is the negative reciprocal of the slope of line segment 𝐴𝐵.

To solve this problem, we’ll need to do three things. First, find the slope of line segment 𝐴𝐵. Use that slope to find the negative reciprocal, which is the slope of line segment 𝐵𝐶. Then, take the slope of line 𝐵𝐶 and use that to find the 𝑦-value in point 𝐶. If we have two points, we find the slope by using 𝑚 equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. For the points 𝐴 and 𝐵, that would be two minus negative one over one minus negative three, which equals three-fourths. The slope of line 𝐴𝐵 is then three-fourths. And we’ve completed step one.

For step two, we need to take the negative reciprocal of the slope we found in step one. The negative reciprocal of three-fourths is negative four-thirds. And that is step two. Now, for step three, we’ll take point 𝐵: one, two and point 𝐶: seven, 𝑦. We’ll let 𝐵 be 𝑥 one, 𝑦 one and 𝐶 be 𝑥 two, 𝑦 two. The slope negative four-thirds is equal to 𝑦 minus two over seven minus one. Seven minus one is six. To solve this, we cross multiply. Negative four times six equals three times 𝑦 minus two. Negative 24 equals three 𝑦 minus six.

To give us a bit more room to solve for 𝑦, we add six to both sides and we get negative 18 equals three 𝑦. Divide both sides of the equation by three and we get negative six equals 𝑦. And we found from step three that 𝑦 must be equal to negative six. This means, for this to be a right triangle, point 𝐶 needs to be located at seven, negative six. And so, we found that missing value to be negative six.

To wrap up, we’ll review our key points. Parallel lines do not intersect and have the same slope. Perpendicular lines intersect at a 90-degree angle and have negative reciprocal slopes of one another.