Topics
10. Connecting Algebra & Geometry with Coordinates
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United States of AmericaFL
Grade 912

10.04 Slopes of parallel and perpendicular lines

Lesson
Practice
Lesson

Concept summary

The slopes of parallel and perpendicular lines have particular relationships.

Slopes of parallel lines theorem

Two non-vertical lines are parallel if and only if their slopes are equal. Any two vertical lines are parallel.

Two lines labeled 1 and 2 on a first and four quadrant coordinate plane without numbers. Line 1 is parallel to line 2. The equation m sub 1 equals m sub 2 is shown.
Slopes of perpendicular lines theorem

Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. A vertical and horizontal line are perpendicular.

Two lines labeled 1 and 2 on a first and four quadrant coordinate plane without numbers. Line 1 is perpendicular to line 2. The equation m sub 1 m sub 2 equals negative 1 is shown.

We may also call the slopes of perpendicular lines opposite reciprocals.

To find the slope of a non-vertical line, we can convert to slope-intercept form, y=mx+b, and identify m.

Worked examples

Example 1

Consider the lines on the given coordinate plane.

-5
-4
-3
-2
-1
1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
4
5
y
a

Identify all pairs of parallel lines.

Approach

We are looking for pairs of lines with the same slope. We can count the rise and run to determine the slope.

Solution

Line a and line d both have a slope of m_a=3=m_d, so they are parallel.

Reflection

Lines b and c have similar slopes, but are not actually parallel.

b

Identify all pairs of perpendicular lines.

Approach

We are looking for pairs of lines with slopes that are opposite reciprocals, so their slopes should have a product of -1.

Looking at the coordinate plane, the pairs that look like they might be perpendicular are a and b, a and c, and e and f.

Solution

From the coordinate plane, we can find that:

m_a=3,\,m_b=-\dfrac{1}{3},\,m_c=-\dfrac{1}{4},\,m_e=\dfrac{1}{2},\,m_f=-2

Line a and line b have slopes with a product of 3 \left( -\dfrac{1}{3}\right) =-1, so they are perpendicular.

Line e and line f have slopes with a product of \dfrac{1}{2} \left( -2\right) =-1, so they are perpendicular.

Reflection

Note that we didn't specify anything about line d. This is because in part (a) we determined that lines a and d are parallel, and so any line that is perpendicular to a is also perpendicular to d, by the perpendicular transversal theorem.

In particular, that means that line b is perpendicular to line d.

Example 2

Consider the line 4x-3y=-6.

a

Find the y-intercept of the line.

Approach

To find the y-intercept, we can either substitute x=0 or rearrange to slope-intercept form.

Solution

Using the strategy of setting x=0,

\displaystyle 4x-3y\displaystyle =\displaystyle -6Given equation
\displaystyle 4\left(0\right)-3y\displaystyle =\displaystyle -6Substitute in x=0
\displaystyle -3y\displaystyle =\displaystyle -6Simplify
\displaystyle y\displaystyle =\displaystyle 2Divide both sides by -3

The y-intercept is \left(0,2\right).

b

Find the equation of the line that is perpendicular to the given line and has the same y-intercept. Write the equation in standard form.

Approach

For the new line to be perpendicular to the given line, its slope must be the opposite reciprocal of the slope of the given line.

Solution

Rearranging the equation of the given line to slope-intercept form gives us:

\displaystyle 4x-3y\displaystyle =\displaystyle -6Given equation
\displaystyle -3y\displaystyle =\displaystyle -4x-6Subtract 4x from both sides
\displaystyle y\displaystyle =\displaystyle \dfrac{4}{3}x+2Divide both sides by -3

The slope of the given line is m=\dfrac{4}{3}.

The slope of a perpendicular line is m_{\perp}=-\dfrac{3}{4}

We know from the previous part that the y-intercept of our new line will be \left( 0,2 \right).

The equation of our new line in slope-intercept form is y=-\dfrac{3}{4}x+2, which we now want to convert to standard form.

\displaystyle y\displaystyle =\displaystyle -\dfrac{3}{4}x+2Equation of new line
\displaystyle 4y\displaystyle =\displaystyle -3x+8Multiply both sides by 4
\displaystyle 3x+4y\displaystyle =\displaystyle 8Add 3x to both sides

The equation of our new line is 3x+4y=8.

Reflection

Converting the equation of the given line to slope-intecept form confirmed our answer to part (a).

Outcomes

MA.912.GR.3.3

Use coordinate geometry to solve mathematical and real-world geometric problems involving lines, circles, triangles and quadrilaterals.

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