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You might assume that you cannot determine the equation of a line from anything other than its slope and y-intercept, but there are several other ways to go about it. For example, perpendicular lines are lines that intersect each other at right angles. Here is an example:

Multiplying the slopes of nearly any two perpendicular lines will give you a product of -1, meaning that they are opposite reciprocals of each other. The lone exceptions are horizontal lines. Since vertical lines have an undefined slope you cannot multiply by that.

Outside of that one exception, we can use the equation of one perpendicular line and a point on the plane to write an equation for the other perpendicular line. This article will show you how to do so.

Let's say we want to write the equation for a line that passes through the point $\left(1,3\right)$ and is perpendicular to the line $y=3x+2$ . We know that multiplying the slope of the equation we have with the equation we need produces a product of -1. Since our equation is in the $y=mx+b$ format, we also know that its slope is 3. This means that we can determine the slope of the equation we need by figuring out what number solves the equation $3{m}_{2}=-1$ . So:

$3{m}_{2}=-1$

$\frac{3{m}_{2}}{3}=-\frac{1}{3}$

${m}_{2}=-\frac{1}{3}$

Now that we know the slope of the line's equation, we can use the point-slope form to write the rest of the equation:

$y-{y}_{1}={m}_{2}(x-{x}_{1})$

Set ${m}_{2}$ to $-\frac{1}{3}$ , x1 to 1 (based on the point we were given), and ${y}_{1}$ to 3 (also based on the coordinates above):

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

Next, apply the Distributive Property to make the equation easier to work with:

$y-3=-\frac{1}{3}x+\frac{1}{3}$

Then, add 3 to each side to isolate the y:

$y=-\frac{1}{3}x+\frac{1}{3}+3$

$y=-\frac{1}{3}x+\frac{10}{3}$

There's our equation! You may be asked to graph both lines when you're finished. Remember to add arrows indicating that both lines extend infinitely in both directions, which would look something like this:

As you can see, we have two perpendicular lines, one of which passes through the point $\left(1,3\right)$ . The most important thing to remember when doing problems like this is to take your time. If you come up with the wrong slope during the first part, it'll be impossible for you to recover and get the right answer at the end.

a. Write an equation for a line that's perpendicular to $x+3y=6$ and passes through the point $\left(1,5\right)$ .

$x+3y=6$

$3y=-x+6$

$y=-\frac{1}{3}x+2$

The slope of the given line is $-\frac{1}{3}$ . The slope of a line perpendicular to this line is the negative reciprocal of $-\frac{1}{3}$ , which is 3. Now, use the point-slope form with the point $\left(1,5\right)$ and the slope 3:

$y-{y}_{1}=m(x-{x}_{1})$

$y-5=3(x-1)$

$y-5=3x-3$

$y=3x+2$

b. Write an equation for a line that's perpendicular to $3x+5y=15$ and passes through the point $\left(3,2\right)$ .

$3x+5y=15$

$5y=-3x+15$

$y=-\frac{3}{5}x+3$

The slope of the given line is $\frac{3}{5}$ . The slope of a line perpendicular to this line is the negative reciprocal of $\frac{3}{5}$ , which is $\frac{5}{3}$ . Now, use the point-slope form with the point $\left(3,2\right)$ and the slope $\frac{5}{3}$ :

$y-{y}_{1}=m(x-{x}_{1})$

$y-2=\frac{5}{3}(x-3)$

$y-2=\frac{5}{3}x-5$

$y=\frac{5}{3}x-3$

c. Write an equation for a line that's perpendicular to $4x-3y=6$ and passes through the point $\left(4,6\right)$ .

$4x-3y=6$

$3y=4x-6$

$y=\frac{4}{3}x-2$

The slope of the given line is $\frac{4}{3}$ . The slope of a line perpendicular to this line is the negative reciprocal of $\frac{4}{3}$ , which is $-\frac{3}{4}$ . Now, use the point-slope form with the point $\left(4,6\right)$ and the slope $-\frac{3}{4}$ :

$y-{y}_{1}=m(x-{x}_{1})$

$y-6=-\frac{3}{4}(x-4)$

$y-6=-\frac{3}{4}x+3$

$y=-\frac{3}{4}x+9$

d. A line has the equation $2y=3x+3$ . What slope would you expect a perpendicular line to have?

$2y=3x+3$

$y=\frac{3}{2}x+\frac{3}{2}$

The slope of the given line is $\frac{3}{2}$ . The slope of a line perpendicular to this line is the negative reciprocal of $\frac{3}{2}$ , which is $-\frac{2}{3}$ .

Perpendicular Transversal Theorem

College Algebra Diagnostic Tests

Using perpendicular lines and slopes to find the equation of a line is almost like a mathematical magic trick, making math more fun to study. However, it also requires students to be comfortable with algebra, graphing, and the Distributive Property. If the student in your life is feeling a little overwhelmed, a private math tutor can provide an environment where it's okay to make mistakes - they're simply part of the educational process. Contact the Educational Directors at Varsity Tutors for more info.

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