Equations of Perpendicular Lines

Hello! Today we’re going to take a look at how to find the equation of a line that passes through a given point and is perpendicular to another line. Let’s start with an example.

Find the equation of a line that passes through the point \((1,7)\) and is perpendicular to the line \(y=4x-3\).

Remember, perpendicular lines have slopes that are negative reciprocals of one another. So, to determine the slope of the line we are looking for, we must first determine the slope of the line we are given. The line \(y=4x-3\) is in slope-intercept form \((y=mx+b)\), so we know that our slope \((m)\) is 4.

\(y=4x-3\)

Now we need to find the negative reciprocal of 4 to determine the slope of our line. The negative reciprocal of 4 is \(-\frac{1}{4}\), so the slope of our line is \(-\frac{1}{4}\).

\(m=-\frac{1}{4}\)

Now that we know the slope of the line and that the line passes through the point \((1,7)\), we can plug these values into either our general slope-intercept equation or our general point-slope equation. For this example, we’ll use the slope-intercept equation.

\(y=mx+b\)

We know our value for \(m\) and values for \(x\) and \(y\) that make our equation true (because the line passes through them), so let’s plug in these three values and solve for \(b\).

\(7=-\frac{1}{4}(1)+b\)

First, multiply.

\(7=-\frac{1}{4}+b\)

Now we need to convert 7 into a fraction that can be added with \(\frac{1}{4}\). So to do that, we need to remember that a whole number is always a fraction over 1. So \(\frac{7}{1}\) is the same as 7. Now, to convert \(\frac{7}{1}\) into a fraction with a denominator of 4, we’ll have to multiply both the denominator and the numerator by 4. That will give us the fraction \(\frac{28}{4}\).

\(\frac{28}{4}=-\frac{1}{4}+b\)

Then, to solve for \(b\), we’re going to add \(\frac{1}{4}\) to both sides of the equation.

\(\frac{29}{4}=b\)

Now that we know what \(b\) is, we can use our values for \(m\) and \(b\) to create the equation for our line.

\(y=-\frac{1}{4}x+\frac{29}{4}\)

Don’t be scared by these fractions! As you continue on to higher math, you will start using fractions more and more frequently, especially improper fractions. They’re actually easier to work with than decimals, so you don’t need to convert these values to decimals, leave them as fractions.

Let’s try another example!

Find the equation of a line that passes through the point \((-2,6)\) and is perpendicular to the line \(y=-2x+3\).

First, find the slope of the line we are given. Again, this is in slope-intercept form, so our slope is \(-2\).

\(y=-2x+3\)

Now, use this value to find the slope of the line we are looking for. Remember, the slopes of perpendicular lines are negative reciprocals of one another, so the slope of the line we are looking for is:

\(m=\frac{1}{2}\)

For this example, we are going to find our equation by plugging in our slope and point into the point-slope equation.

\(y-y_{1}=m(x-x_{1})\)

Remember, since this form is created to tell you a specific point on the line, we will plug in our point \((-2,6)\) for \((x_{1},y_{1})\) instead of \((x,y)\) like we did last time. Let’s plug in what we know.

\(y-6=\frac{1}{2}(x-(-2))\)

Now, if we want our answer in point-slope form, we can stop right here. However, if we want to have it in slope-intercept form, we need to do a little bit of manipulating. First, change the subtracting a negative number to addition.

\(y-6=\frac{1}{2}(x+2)\)

Now, distribute the \(\frac{1}{2}\) on the right side of the equation.

\(y-6=\frac{1}{2}x+1\)

Then, add 6 to both sides.

\(y-6+6=\frac{1}{2}x+1+6\)

\(y=\frac{1}{2}x+7\)

And there you have it! Before we go, let’s try one more problem together.

Find the equation of a line that passes through the point \((5,-7)\) and is perpendicular to the line \(y=-\frac{1}{3}x-9\).

First, find the slope of the given line. This line is in slope-intercept form, so the slope is \(-\frac{1}{3}\).

\(y=-\frac{1}{3}x-9\)

Now, find the opposite reciprocal of this number to find the slope of the perpendicular line.

\(m=3\)

Then, plug in your slope and point into either the slope-intercept equation or the point-slope equation. We’ll work through both ways for this example, starting with the slope-intercept equation.

\(y=mx+b\)

Plug in the values for \(m\), \(x\), and \(y\). Then, solve for \(b\).

\(-7=3(5)+b\)
 
\(-7=15+b\)
 
\(-22=b\)

Now that we know our value for \(b\), we can plug it into the slope-intercept equation, along with the value for \(m\).

\(y=3x-22\)

If we wanted to use the point-slope equation instead, we would get the same answer. Let’s try it that way!

\(y-y_{1}=m(x-x_{1})\)

Plug in the values for \(m\), \(x_{1}\), and \(y_{1}\).

\(y-(-7)=3(x-5)\)

If we simplify the subtracting a negative number part, we get:

\(y+7=3(x-5)\)

Now, we can distribute and rearrange to convert our equation to slope-intercept form.

\(y+7=3x-15\)

Subtract 7 from both sides of the equation.

\(y=7-7=3x-15-7\)
 
\(y=3x-22\)

Our answers match, just like they should! This proves that either way you want to solve the equation, using slope-intercept form or point-slope form, will get you the correct answer.

I hope that this video on equations of perpendicular lines was helpful. Thanks for watching, and happy studying!

Practice Questions