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# 3.5: Parallel and Perpendicular Lines in the Coordinate Plane

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Compute slope.
• Determine the equation of parallel and perpendicular lines.

## Review Queue

Find the slope between the following points.

1. (-3, 5) and (2, -5)
2. (7, -1) and (-2, 2)
3. Is \begin{align*}x = 3\end{align*} horizontal or vertical? How do you know?
4. Is \begin{align*}y = -1\end{align*} horizontal or vertical? How do you know?
5. Graph \begin{align*}y=\frac{1}{4}x-2\end{align*} on an \begin{align*}x-y\end{align*} plane.

Know What? The picture to the right is the California Incline, a short road that connects Highway 1 with Santa Monica. The length of the road is 1532 feet and has an elevation of 177 feet. You may assume that the base of this incline is zero feet. Can you find the slope of the California Incline?

HINT: You will need to use the Pythagorean Theorem, which you may have seen in a previous math class.

## Slope in the Coordinate Plane

Recall from Algebra I, two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*} have a slope of \begin{align*}m = \frac{(y_2-y_1)}{(x_2-x_1)}\end{align*}.

Different Types of Slope:

Positive

Negative

Zero

Undefined

Example 1: What is the slope of the line through (2, 2) and (4, 6)?

Solution: Use (2, 2) as \begin{align*}(x_1, y_1)\end{align*} and (4, 6) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m=\frac{6-2}{4-2} = \frac{4}{2} = 2\end{align*}

This slope is positive. Slope can also be written “rise over run.” In this case we “rise” 2, and “run” in the positive direction 1.

Example 2: Find the slope between (-8, 3) and (2, -2).

Solution: Use (-8, 3) as \begin{align*}(x_1, y_1)\end{align*} and (2, -2) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m = \frac{-2-3}{2-(-8)} = \frac{-5}{10} = -\frac{1}{2}\end{align*}

This slope is negative. Here, we don’t “rise,” but “fall” one, and “run” in the positive direction 2.

Example 3: Find the slope between (-5, -1) and (3, -1).

Solution: Use (-5, -1) as \begin{align*}(x_1, y_1)\end{align*} and (3, -1) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m=\frac{-1-(-1)}{3-(-5)} = \frac{0}{8} = 0\end{align*}

The slope of this line is 0, or a horizontal line. Horizontal lines always pass through the \begin{align*}y-\end{align*}axis. The \begin{align*}y-\end{align*}coordinate for both points is -1.

So, the equation of this line is \begin{align*}y = -1\end{align*}.

Example 4: What is the slope of the line through (3, 2) and (3, 6)?

Solution: Use (3, 2) as \begin{align*}(x_1, y_1)\end{align*} and (3, 6) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m = \frac{6-2}{3-3} = \frac{4}{0} = undefined\end{align*}

The slope of this line is undefined, which means that it is a vertical line. Vertical lines always pass through the \begin{align*}x-\end{align*}axis. The \begin{align*}x-\end{align*}coordinate for both points is 3.

So, the equation of this line is \begin{align*}x = 3\end{align*}.

## Slopes of Parallel Lines

Earlier in the chapter we defined parallel lines as two lines that never intersect. In the coordinate plane, that would look like this:

If we take a closer look at these two lines, the slopes are both \begin{align*}\frac{2}{3}\end{align*}.

This can be generalized to any pair of parallel lines.

Parallel lines have the same slope.

Example 5: Find the equation of the line that is parallel to \begin{align*}y=-\frac{1}{3}x+4\end{align*} and passes through (9, -5).

Solution: Recall that the equation of a line is \begin{align*}y = mx + b\end{align*}, where \begin{align*}m\end{align*} is the slope and \begin{align*}b\end{align*} is the \begin{align*}y-\end{align*}intercept. We know that parallel lines have the same slope, so the line will have a slope of \begin{align*}-\frac{1}{3}\end{align*}. Now, we need to find the \begin{align*}y-\end{align*}intercept. Plug in 9 for \begin{align*}x\end{align*} and -5 for \begin{align*}y\end{align*} to solve for the new \begin{align*}y-\end{align*}intercept \begin{align*}(b)\end{align*}.

\begin{align*}-5 & = -\frac{1}{3}(9)+b\\ -5 & = -3 + b\\ -2 & = b\end{align*}

The equation of line is \begin{align*}y = -\frac{1}{3}x-2\end{align*}.

Parallel lines always have the same slope and different \begin{align*}y-\end{align*}intercepts.

## Slopes of Perpendicular Lines

Perpendicular lines are two lines that intersect at a \begin{align*}90^\circ\end{align*}, or right, angle. In the coordinate plane, that would look like this:

If we take a closer look at these two lines, the slope of one is -4 and the other is \begin{align*}\frac{1}{4}\end{align*}.

This can be generalized to any pair of perpendicular lines in the coordinate plane.

The slopes of perpendicular lines are opposite signs and reciprocals of each other.

Example 6: Find the slope of the perpendicular lines to the lines below.

(a) \begin{align*}y=2x+3\end{align*}

(b) \begin{align*}y = -\frac{2}{3}x-5\end{align*}

(C) \begin{align*}y=x+2\end{align*}

Solution: Look at the slope of each of these.

(a) \begin{align*}m = 2\end{align*}, so \begin{align*}m_\perp\end{align*} is the reciprocal and negative, \begin{align*}m_\perp = -\frac{1}{2}\end{align*}.

(b) \begin{align*}m = -\frac{2}{3}\end{align*}, take the reciprocal and make the slope positive, \begin{align*}m_\perp=\frac{2}{3}\end{align*}.

(c) Because there is no number in front of \begin{align*}x\end{align*}, the slope is 1. The reciprocal of 1 is 1, so the only thing to do is make it negative, \begin{align*}m_\perp= -1\end{align*}.

Example 7: Find the equation of the line that is perpendicular to \begin{align*}y=-\frac{1}{3}x+4\end{align*} and passes through (9, -5).

Solution: First, the slope is the reciprocal and opposite sign of \begin{align*}-\frac{1}{3}\end{align*}. So, \begin{align*}m = 3\end{align*}. Plug in 9 for \begin{align*}x\end{align*} and -5 for \begin{align*}y\end{align*} to solve for the new \begin{align*}y-\end{align*}intercept \begin{align*}(b)\end{align*}.

\begin{align*}-5 & = 3(9)+b\\ -5 & = 27+b\\ -32 & = b\end{align*}

Therefore, the equation of line is \begin{align*}y=3x-32\end{align*}.

## Graphing Parallel and Perpendicular Lines

Example 8: Find the equation of the lines below and determine if they are parallel, perpendicular or neither.

Solution: The top line has a \begin{align*}y-\end{align*}intercept of 1. From there, use “rise over run” to find the slope. From the \begin{align*}y-\end{align*}intercept, if you go up 1 and over 2, you hit the line again, \begin{align*}m = \frac{1}{2}\end{align*}. The equation is \begin{align*}y=\frac{1}{2}x+1\end{align*}.

For the second line, the \begin{align*}y-\end{align*}intercept is -3. The “rise” is 1 and the “run” is 2 making the slope \begin{align*}\frac{1}{2}\end{align*}. The equation of this line is \begin{align*}y=\frac{1}{2}x-3\end{align*}.

The lines are parallel because they have the same slope.

Example 9: Graph \begin{align*}3x-4y=8\end{align*} and \begin{align*}4x+3y=15\end{align*}. Determine if they are parallel, perpendicular, or neither.

Solution: First, we have to change each equation into slope-intercept form. In other words, we need to solve each equation for \begin{align*}y\end{align*}.

\begin{align*}3x-4y & = 8 && 4x+3y = 15\\ -4y & = -3x+8 && 3y = -4x + 15\\ y & = \frac{3}{4}x-2 && y = -\frac{4}{3}x+5\end{align*}

Now that the lines are in slope-intercept form (also called \begin{align*}y-\end{align*}intercept form), we can tell they are perpendicular because the slopes are opposites signs and reciprocals.

Example 10: Find the equation of the line that is

(a) parallel to the line through the point.

(b) perpendicular to the line through the points.

Solution: First the equation of the line is \begin{align*}y=2x+6\end{align*} and the point is (2, -2). The parallel would have the same slope and pass through (2, -2).

\begin{align*}y & =2x+b\\ -2 & = 2(2) + b\\ -2 & = 4+b\\ -6 & = b\end{align*}

The equation is \begin{align*}y=2x+-6\end{align*}

The perpendicular line also goes through (2, -2), but the slope is \begin{align*}-\frac{1}{2}\end{align*}.

\begin{align*}y & = -\frac{1}{2}x+b\\ -2 & = -\frac{1}{2}(2) + b\\ -2 & = -1+b\\ -1 & =b\end{align*}

The equation is \begin{align*}y = -\frac{1}{2}x-1\end{align*}

Know What? Revisited In order to find the slope, we need to first find the horizontal distance in the triangle to the right. This triangle represents the incline and the elevation. To find the horizontal distance, we need to use the Pythagorean Theorem, \begin{align*}a^2+b^2 = c^2\end{align*}, where \begin{align*}c\end{align*} is the hypotenuse.

\begin{align*}177^2 + run^2 & = 1532^2\\ 31,329 + run^2 &= 2,347,024\\ run^2 & = 2,315,695\\ run & \approx 1521.75\end{align*}

The slope is then \begin{align*}\frac{177}{1521.75}\end{align*}, which is roughly \begin{align*}\frac{3}{25}\end{align*}.

## Review Questions

• Questions 1-6 are similar to Examples 1, 2, 3, and 4.
• Questions 7-14 are similar to Example 6 and 9.
• Questions 15-18 are similar to Example 5.
• Questions 19-22 are similar to Example 7.
• Questions 23-26 are similar to Example 8.
• Questions 27-30 are similar to Example 10.

Find the slope between the two given points.

1. (4, -1) and (-2, -3)
2. (-9, 5) and (-6, 2)
3. (7, 2) and (-7, -2)
4. (-6, 0) and (-1, -10)
5. (1, -2) and (3, 6)
6. (-4, 5) and (-4, -3)

Determine if each pair of lines are parallel, perpendicular, or neither. Then, graph each pair on the same set of axes.

1. \begin{align*}y=-2x+3\end{align*} and \begin{align*}y = \frac{1}{2}x+3\end{align*}
2. \begin{align*}y=4x-2\end{align*} and \begin{align*}y=4x+5\end{align*}
3. \begin{align*}y=-x+5\end{align*} and \begin{align*}y=x+1\end{align*}
4. \begin{align*}y=-3x+1\end{align*} and \begin{align*}y=3x-1\end{align*}
5. \begin{align*}2x-3y=6\end{align*} and \begin{align*}3x+2y=6\end{align*}
6. \begin{align*}5x+2y=-4\end{align*} and \begin{align*}5x+2y=8\end{align*}
7. \begin{align*}x-3y=-3\end{align*} and \begin{align*}x+3y=9\end{align*}
8. \begin{align*}x+y=6\end{align*} and \begin{align*}4x+4y=-16\end{align*}

Determine the equation of the line that is parallel to the given line, through the given point.

1. \begin{align*}y=-5x+1; \ (-2, 3)\end{align*}
2. \begin{align*}y=\frac{2}{3}x-2; \ (9, 1)\end{align*}
3. \begin{align*}x-4y=12; \ (-16, -2)\end{align*}
4. \begin{align*}3x+2y=10; \ (8, -11)\end{align*}

Determine the equation of the line that is perpendicular to the given line, through the given point.

1. \begin{align*}y=x-1; \ (-6, 2)\end{align*}
2. \begin{align*}y=3x+4; \ (9, -7)\end{align*}
3. \begin{align*}5x-2y= 6; \ (5, 5)\end{align*}
4. \begin{align*}y = 4; \ (-1, 3)\end{align*}

Find the equation of the two lines in each graph below. Then, determine if the two lines are parallel, perpendicular or neither.

For the line and point below, find:

(a) A parallel line, through the given point.

(b) A perpendicular line, through the given point.

1. \begin{align*}m = \frac{-5-5}{2 + 3} = \frac{-10}{2} = -5\end{align*}
2. \begin{align*}m = \frac{2 + 1}{-2-7} = \frac{3}{-9} = -\frac{1}{3}\end{align*}
3. Vertical because it has to pass through \begin{align*}x = 3\end{align*} on the \begin{align*}x-\end{align*}axis and doesn’t pass through \begin{align*}y-\end{align*}axis at all.
4. Horizontal because it has to pass through \begin{align*}y = -1\end{align*} on the \begin{align*}y-\end{align*}axis and it does not pass through the \begin{align*}x-\end{align*}axis at all.

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