In most cities, it is very common for the streets to be laid out in a grid format. Below is an example. (Assume the lower left corner is the origin.) Are any of the streets perpendicular? What are the slopes of each street?

### Guidance

When two lines are perpendicular, they intersect at a \begin{align*}90^\circ,\end{align*} or right, angle. The slopes of two perpendicular lines, are therefore, not the same. Let’s investigate the relationship of perpendicular lines.

#### Investigation: Slopes of Perpendicular Lines

Tools Needed: Pencil, ruler, protractor, and graph paper

- Draw a @$\begin{align*}x-y\end{align*}@$ plane that goes from -5 to 5 in both the @$\begin{align*}x\end{align*}@$ and @$\begin{align*}y\end{align*}@$ directions.
- Plot (0, 0) and (1, 3). Connect these to form a line.
- Plot (0, 0) and (-3, 1). Connect these to form a second line.
- Using a protractor, measure the angle formed by the two lines. What is it?
- Use slope triangles to find the slope of both lines. What are they?
- Multiply the slope of the first line times the slope of the second line. What do you get?

From this investigation, the lines from #2 and #3 are perpendicular because they form a @$\begin{align*}90^\circ\end{align*}@$ angle. The slopes are 3 and @$\begin{align*}- \frac{1}{3},\end{align*}@$ respectively. When multiplied together, the product is -1. This is true of all perpendicular lines.

*The product of the slopes of two perpendicular lines is -1.*

*If a line has a slope of @$\begin{align*}m,\end{align*}@$ then the perpendicular slope is @$\begin{align*}- \frac{1}{m}\end{align*}@$.*

#### Example A

Find the equation of the line that is perpendicular to @$\begin{align*}2x -3y = 15\end{align*}@$ and passes through (6, 5).

**Solution:** First, we need to change the line from standard to slope-intercept form.

@$$\begin{align*}2x - 3y &= 15\\ -3y &=-2x + 15\\ y &= \frac{2}{3}x - 5\end{align*}@$$

Now, let’s find the perpendicular slope. From the investigation above, we know that the slopes must multiply together to equal -1.

@$$\begin{align*}\frac{2}{3} \cdot m &= -1\\ \xcancel{\frac{3}{2} \cdot \frac{2}{3}} \cdot m &= -1 \cdot \frac{3}{2}\\ m &= - \frac{3}{2}\end{align*}@$$

Notice that the perpendicular slope is the *opposite sign and reciprocals* with the original slope. Now, we need to use the given point to find the @$\begin{align*}y-\end{align*}@$intercept.

@$$\begin{align*}5 &= - \frac{3}{2}(6) + b\\ 5 &= -9 + b\\ 14 &= b\end{align*}@$$

The equation of the line that is perpendicular to @$\begin{align*}y = \frac{2}{3}x - 5 \end{align*}@$ is @$\begin{align*}y = - \frac{3}{2}x + 14\end{align*}@$.

If we write these lines in standard form, the equations would be @$\begin{align*}2x - 3y = 15\end{align*}@$ and @$\begin{align*}3x + 2y = 28,\end{align*}@$ respectively.

#### Example B

Write the equation of the line that passes through (4, -7) and is perpendicular to @$\begin{align*}y = 2\end{align*}@$.

**Solution:** The line @$\begin{align*}y = 2\end{align*}@$ does not have a @$\begin{align*}x-\end{align*}@$term, meaning it has no slope and a horizontal line. Therefore, to find the perpendicular line that passes through (4, -7), it would have to be a vertical line. Only need the @$\begin{align*}x-\end{align*}@$coordinate. The perpendicular line would be @$\begin{align*}x = 4\end{align*}@$.

#### Example C

Write the equation of the line that passes through (6, -10) and is perpendicular to the line that passes through (4, -6) and (3, -4).

**Solution:** First, we need to find the slope of the line that our line will be perpendicular to. Use the points (4, -6) and (3, -4) to find the slope.

@$$\begin{align*}m = \frac{-4-(-6)}{3-4} = \frac{2}{-1} = -2\end{align*}@$$

Therefore, the perpendicular slope is the opposite sign and the reciprocal of -2. That makes the slope @$\begin{align*}\frac{1}{2}\end{align*}@$. Use the point (6, -10) to find the @$\begin{align*}y-\end{align*}@$intercept.

@$$\begin{align*}-10 &= \frac{1}{2}(6) + b\\ -10 &= 3 + b \\ -7 &= b\end{align*}@$$

The equation of the perpendicular line is @$\begin{align*}y = \frac{1}{2}x - 7\end{align*}@$.

**Intro Problem Revisit** Using the graphed endpoints of each street, we find the slope of A St. is @$\begin{align*}- \frac{1}{2}\end{align*}@$, 7th Ave. is @$\begin{align*}2\end{align*}@$, and the slope is Lincoln Blvd. is @$\begin{align*}\frac{1}{2}\end{align*}@$. Therefore, because they are negative reciprocals of each other, A St. and 7th Ave. are perpendicular. Lincoln Blvd. is neither perpendicular nor parallel to 7th Ave. or A Street.

### Guided Practice

1. Find the equation of the line that is perpendicular to @$\begin{align*}x - 2y = 8\end{align*}@$ and passes through (4, -3).

2. Find the equation of the line that passes through (-8, 7) and is perpendicular to the line that passes through (6, -1) and (1, 3).

3. Are @$\begin{align*}x - 4y = 8\end{align*}@$ and @$\begin{align*}2x + 8y = -32\end{align*}@$ parallel, perpendicular or neither?

#### Answers

1. First, we need to change this line from standard form to slope-intercept form.

@$$\begin{align*}x -2y &= 8\\ -2y &= -x + 8\\ y &=\frac{1}{2}x - 4\end{align*}@$$

The perpendicular slope will be -2. Let's find the new @$\begin{align*}y-\end{align*}@$intercept.

@$$\begin{align*}-3 &= -2(4) + b\\ -3 &= -8 + b \\ 5 &= b\end{align*}@$$

The equation of the perpendicular line is @$\begin{align*}y = -2x + 5\end{align*}@$ or @$\begin{align*}2x + y = 5\end{align*}@$.

2. First, find the slope between (6, -1) and (1, 3).

@$$\begin{align*}m = \frac{-1-3}{6-1} = \frac{-4}{5} = -\frac{4}{5}\end{align*}@$$

From this, the slope of the perpendicular line will be @$\begin{align*}\frac{5}{4}\end{align*}@$. Now, use (-8, 7) to find the @$\begin{align*}y-\end{align*}@$intercept.

@$$\begin{align*}7 &= \frac{5}{4}(-8) + b\\ 7 &= -10 + b \\ 17 &= b\end{align*}@$$

The equation of the perpendicular line is @$\begin{align*}y = \frac{5}{4}x + 17\end{align*}@$.

3. To determine if the two lines are parallel or perpendicular, we need to change them both into slope-intercept form.

@$$\begin{align*}x - 4y &= 8 \qquad \qquad \qquad 2x + 8y = -32\\ -4y &= -x + 8 \quad and \qquad \quad \ 8y = -2x -32\\ y &=\frac{1}{4}x - 2 \qquad \qquad \qquad \ y = - \frac{1}{4}x - 4\end{align*}@$$

Now, just look at the slopes. One is @$\begin{align*}\frac{1}{4}\end{align*}@$ and the other is @$\begin{align*}- \frac{1}{4}\end{align*}@$. They are not the same, so they are not parallel. To be perpendicular, the slopes need to be reciprocals, which they are not. Therefore, these two lines are not parallel or perpendicular.

### Explore More

Find the equation of the line, given the following information. You may leave your answer in slope-intercept form.

- Passes through (4, 7) and is perpendicular to @$\begin{align*}x - y = -5\end{align*}@$.
- Passes through (-6, -2) and is perpendicular to @$\begin{align*}y = 4\end{align*}@$.
- Passes through (4, 5) and is perpendicular to @$\begin{align*}y = - \frac{1}{3}x - 1\end{align*}@$.
- Passes through (1, -9) and is perpendicular to @$\begin{align*}x = 8\end{align*}@$.
- Passes through (0, 6) and perpendicular to @$\begin{align*}x - 4y = 10\end{align*}@$.
- Passes through (-12, 4) and is perpendicular to @$\begin{align*}y = -3x + 5\end{align*}@$.
- Passes through the @$\begin{align*}x-\end{align*}@$intercept of @$\begin{align*}2x - 3y = 6\end{align*}@$ and perpendicular to @$\begin{align*}x + 6y = -3\end{align*}@$.
- Passes through (7, -8) and is perpendicular to @$\begin{align*}2x + 5y = 14\end{align*}@$.
- Passes through (1, 3) and is perpendicular to the line that passes through (-6, 2) and (-4, 6).
- Passes through (3, -10) and is perpendicular to the line that passes through (-2, 2) and (-8, 1).
- Passes through (-4, -1) and is perpendicular to the line that passes through (-15, 7) and (-3, 3).

Are the pairs of lines parallel, perpendicular or neither?

- @$\begin{align*}4x + 2y = 5\end{align*}@$ and @$\begin{align*}5x - 10y = -20\end{align*}@$
- @$\begin{align*}9x + 12y = 8\end{align*}@$ and @$\begin{align*}6x + 8y = -1\end{align*}@$
- @$\begin{align*}5x - 5y = 20\end{align*}@$ and @$\begin{align*}x + y = 7\end{align*}@$
- @$\begin{align*}8x -4y = 12\end{align*}@$ and @$\begin{align*}4x - y = -15\end{align*}@$