Topics
10. Connecting Algebra and Geometry through Coordinates
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United States of America
Grades 9-12

10.03 Parallel and perpendicular lines

Lesson
Practice

Introduction

We will continue building our understanding of parallel and perpendicular lines by considering characterstics seen in their rates of change. We will use the slope formula from Algebra 1 lesson  3.05 Graphing linear functions  to prove whether lines are parallel or perpendicular in the coordinate plane.

Parallel lines

Exploration

Move each of the blue points around. The points on the y-axis will move up and down, and the other point can move anywhere.

  1. What do you notice about the slopes of the two parallel lines?
  2. Do you think this is true for all non-vertical lines?
  3. How could we prove two lines are parallel?
Loading interactive...

Parallel lines will always have the same slope. This means they will never intersect. Any two vertical lines are parallel.

Slopes of parallel lines theorem

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

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.

We can use this theorem to prove lines are parallel and to find the equations of parallel lines.

Examples

Example 1

Prove that the lines are parallel.

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Worked Solution
Create a strategy

We will use the slopes of parallel lines theorem to prove the lines are parallel by showing the slopes are equal.

If we find nice points, we can count the rise and run to determine the slope.

Apply the idea
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Counting the rise and run, we see the top line has a slope of -\dfrac{1}{6}. Doing the same for the bottom line, the slope is also -\dfrac{1}{6}.

Therefore, m_{1}=m_{2}=-\dfrac{1}{6} so these lines are parallel by the slopes of parallel lines theorem.

Example 2

The line AB passes through the points \left(-2,9\right) and \left(3,-21\right).

a

Write the equation of the line.

Worked Solution
Create a strategy

To find the equation, we need to know the slope and the y-intercept. We will find the slope using m=\dfrac{y_2-y_1}{x_2-x_1}, and we will find the y-intercept using y=mx+b.

Apply the idea
\displaystyle m\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle =\displaystyle \dfrac{-21-9}{3-(-2)}Substitute (x_1,y_1) and (x_2,y_2)
\displaystyle =\displaystyle \dfrac{-30}{5}Evaluate the subtraction
\displaystyle =\displaystyle -6Evaluate the division

The slope of the line is m=-6. Now, we will use y=mx+b with the slope we found and one of the points. We can use either point because either will result in the same answer.

\displaystyle y\displaystyle =\displaystyle mx+bSlope-intercept form of a linear equation
\displaystyle 9\displaystyle =\displaystyle -6(-2)+bSubstitute m=-6 and (x_1,y_1)
\displaystyle 9\displaystyle =\displaystyle 12+bEvaluate the multiplication
\displaystyle -3\displaystyle =\displaystyle bSubtraction property of equality
\displaystyle b\displaystyle =\displaystyle -3Reflexive property of equality

This means the y-intercept is at \left(0,-3\right).

Substituting m=-6 and b=-3 into slope-intercept form of a linear equation, we find the equation of the line to be y=-6x-3.

Reflect and check

The equation of the line in standard form is 6x+y=-3.

b

Find the equation of the line that passes through \left(1,5\right) and is parallel to the line AB.

Worked Solution
Create a strategy

Since this line is parallel to the line AB, we know that it will have the same slope as \overleftrightarrow{AB} which was -6. We only need to find the y-intercept of the parallel line.

Apply the idea

Just like we did in the previous part, we will substitute the slope and the x- and y-values of the point into y=mx+b.

\displaystyle y\displaystyle =\displaystyle mx+bSlope-intercept form of a linear equation
\displaystyle 5\displaystyle =\displaystyle -6(1)+bSubstitute m=-6 and the point (1,5)
\displaystyle 5\displaystyle =\displaystyle -6+bEvaluate the multiplication
\displaystyle 11\displaystyle =\displaystyle bAddition property of equality
\displaystyle b\displaystyle =\displaystyle 11Reflexive property of equality

The equation of the parallel line is y=-6x+11.

Reflect and check

Using technology to graph the lines, we can see that they are parallel, and they pass through the specified points from parts (a) and (b).

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Example 3

Prove that two non-vertical lines are parallel if their slopes are the same.

Worked Solution
Create a strategy

Begin with the definition of parallel lines: "Parallel lines are lines in the same plane that do not intersect."

Apply the idea

We can prove this informally using definitions and diagrams. We will start with one line on the coordinate plane and try to find a line which never intersects it.

For a line to never intersect the one that we started with, it must be a translation of our starting line without any reflection or rotation. A translation is defined as "a transformation in which every point in a figure is moved in the same direction and by the same distance."

This diagram shows the possible vertical translations.

Whether we translate the line up or down, the slope triangle is translated along with the line. Translations preserve distance, so the distances a and b are preserved.

This means the rise and run remain the same, so the slope of the line remains the same.

This diagram shows the possible horizontal translations. Whether we translate the line left or right, the slope triangle is translated with the line.

Using the same reasoning as we did above, we know the slope of the line remains the same.

Since translating a line does not change its slope, we know that two lines are parallel if and only if their slopes are the same.

Reflect and check

Another strategy for proving this is to make a system of two non-vertical linear equations which has no solution. We can start with:

\begin{cases} y=m_1x+b_1 \\ y=m_2x+b_2 \end{cases}

If b_1=b_2, then system will have a solution when x=0, so we must have b_1\neq b_2.

Now, we can consider what a solution to the equation will look like. If we solve the equation by letting the y-values be equal, we get:

\displaystyle m_1x+b_1\displaystyle =\displaystyle m_2x+b_2Equate y-values
\displaystyle m_1x-m_2x\displaystyle =\displaystyle b_1+b_2Add b_1 and subtract m_2x from both sides
\displaystyle \left(m_1-m_2\right)x\displaystyle =\displaystyle b_1+b_2Factor out x
\displaystyle x\displaystyle =\displaystyle \dfrac{b_1+b_2}{m_1-m_2}Divide both sides by m_1-m_2

We can see that there will be no viable solutions to the system only when m_1=m_2 because that would force us to divide by zero which is not possible. So two lines are parallel if and only if their slopes are the same.

Idea summary

If two lines are parallel, they have the same slope.

If two lines have the same slope, they are parallel.

Perpendicular lines

Exploration

Drag the blue points around. Keep the point on the red line in between the other two points for easier usability.

  1. What do you notice about the slopes of the two perpendicular lines?
  2. Do you think this is true for all non-vertical perpendicular lines?
  3. How could we prove two lines are perpendicular?
Loading interactive...

Perpendicular lines have slopes with opposite signs and they are reciprocals of one another. A vertical and horizontal line are perpendicular.

Slopes of perpendicular lines theorem

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

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 which refers to them being reciprocals with opposite signs. We use this theorem to prove lines are perpendicular and to find the equations of perpendicular lines.

Examples

Example 4

Consider the lines on the given coordinate plane.

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a

Identify all pairs of parallel lines.

Worked Solution
Create a strategy

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

Apply the idea

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

Reflect and check

Lines b and c have similar slopes, but they are not equal.

m_b=-\dfrac{1}{3}

m_c=-\dfrac{1}{4}

b

Identify all pairs of perpendicular lines.

Worked Solution
Create a strategy

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.

Apply the idea

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.

In part (a), we determined that lines a and d are parallel, 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.

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

Example 5

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

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

Worked Solution
Create a strategy

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.

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

Apply the idea

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 equation that the y-intercept of this line is \left( 0,2 \right), and our new line will have the same y-intercept.

The equation of our new line in slope-intercept form is y=-\dfrac{3}{4}x+2.

Reflect and check

Graphing the lines on the same coordinate plane, we can see that there is a 90\degree angle where the lines intersect, and they have the same y-intercept.

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b

Write the equation in standard form.

Worked Solution
Create a strategy

Standard form is when the x- and y-terms are on the same side of the equation, and the coefficients are all integers.

Apply the idea
\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 the new line in standard form is 3x+4y=8.

Reflect and check

Comparing the two equations in standard form,

3x+4y=8

4x-3y=-6

we see that the coefficients of x and y are switched, and one of the signs is opposite. This is what gives us the reciprocal slopes with opposite signs.

Example 6

A mirror is placed along the x-axis. A laser beam is projected along the line y=-x+4 which reflects off the mirror.

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A normal is a line which is perpendicular to the surface of the mirror at the point of reflection. Find the equation of the normal.

Worked Solution
Create a strategy

We can do a quick sketch of the normal to help:

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Apply the idea

Since the mirror is a horizontal line, the normal must be a vertical line if it is to be perpendicular. This means it will be of the form x=a.

Since it goes through the point where the laser hits the mirror, \left(4,0\right), the equation of the normal will be x=4.

b

The angles that the laser and its reflection make with the normal will be congruent. If the angle between the laser beam and the normal is 45 \degree, find the equation of the path of the reflection.

Worked Solution
Create a strategy

Since the angles are congruent, know that the angle formed between the normal and the reflection will also be 45 \degree.

We can label this on our diagram:

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Apply the idea

Using the angle addition postulate, we can show that the angle formed between the laser and its reflection will be 45 \degree+45 \degree=90\degree.

This means that the laser beam path and the reflection path are perpendicular, so their slopes will be negative reciprocals. Since the slope of the laser beam is -1, this means the slope of the reflection will be 1.

The reflection starts at the point \left(4,0\right). Putting all of this together we get:

\displaystyle y\displaystyle =\displaystyle mx+bEquation of a line in slope-intercept form
\displaystyle y\displaystyle =\displaystyle 1x+bSubstitute m_{\perp}=1
\displaystyle 0\displaystyle =\displaystyle 1(4)+bSubstitute \left(4,0\right)
\displaystyle -4\displaystyle =\displaystyle bSubtract 4 from both sides
\displaystyle y\displaystyle =\displaystyle x-4Using m=1 and b=-4

The equation of the reflection is y=x-4.

Reflect and check

We did not need to be told that the angle formed between the laser beam and the normal was 45 \degree. Since the slope of the line was -1, we could form a right isosceles triangle with legs of length 4, so the angle must be 45 \degree.

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Example 7

Prove that two non-vertical lines are perpendicular if the product of their slopes is -1.

Worked Solution
Create a strategy

Begin with the definition of perpendicular lines: "Perpendicular lines are two lines that are at right angles to each other."

Apply the idea

One way to show they are at right angles is to start with one line and consider its slope using a right triangle. For example:

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To find a line that is at a right angle to our starting line, we can just rotate it by 90\degree.

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Notice that rotating the line by 90\degree also rotated the right triangle by 90\degree.

This shows us that the values of the rise and run have switched, so the slope of the new line will be the reciprocal of the starting line.

We can also see that the slope has changed sign, since rotating any line by 90\degree will change its slope from positive to negative or vice versa. The sign of the slope of the new line will be the negative of the starting line.

Therefore, two lines are perpendicular if their slopes are opposite reciprocals, which is the same as their product being -1.

Reflect and check

To show that two non-vertical lines whose slopes have a product of -1 are perpendicular, we would need to begin with a line that has a slope of \dfrac{a}{b} and a line that has a slope of -\dfrac{b}{a} and show they meet at a 90\degree angle.

Idea summary

The slopes of perpendicular lines are reciprocals with opposite signs. When multiplied together, they have a product of -1.

Outcomes

G.CO.D.12

Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

G.GPE.B.5

Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems.

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