Topics
3. Linear Functions
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Algebra 1

3.06 Equations of parallel and perpendicular lines

Lesson
Practice

Parallel and perpendicular lines

Exploration

Click and drag the points on the red line. The points on the y-axis will only move up and down, and the other point can move anywhere.

The blue line can be changed by editing the slope and y-intercept in the boxes below.

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  1. Move the red line until you get the message "a and b are parallel lines". What do you notice about the slopes when the two lines are parallel?
  2. Can you create two different parallel lines with the same y-intercept?
  3. What do you notice about the slopes when the two lines are perpendicular? Move the red line until you get the message "a and b are perpendicular lines".
  4. Can you create two distinct perpedicular lines with the same y-intercept?
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Parallel lines will always have the same slope. This means they will never intersect.

If two parallel lines have the same y-intercept, they will be exactly the same line.

To write the equation for any parallel line, we only need a point on the line. Consider a line parallel to y = 3x-2 that goes through the point (3,\,-2).

Using the point-slope formula, we have:

y - (-2) = 3(x - 3)

y +2 = 3(x - 3)

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Perpendicular lines have slopes with opposite signs and they are reciprocals of one another. These are called negative reciprocals (or opposite reciprocals).

The product of slopes of perpendicular lines is -1 unless one of the lines has an undefined slope.

The lines y = 3x-2 and y = -\dfrac{1}{3}x+4 are perpendicular.

Perpendicular lines may have the same y-intercept.

Consider a line perpendicular to y = 3x-2 with the same y-intercept. We know the slope must be the negative reciprocal which is m = -\dfrac{1}{3} and the y-intercept must be the same, which is b = -2.

Therefore, the equation of a perpendicular line with the same y-intercept is y = -\dfrac{1}{3}x - 2.

Examples

Example 1

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 2

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

a

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 oppposite signs.

c

Determine the zeros of the linear function.

Worked Solution
Create a strategy

Zeros of a function occur at the x-intercepts when y=0.

Apply the idea

In standard form from part (b), we have the equation 3x+4y = 8.

\displaystyle 3x+4y\displaystyle =\displaystyle 8Original equation
\displaystyle 3x+4(0)\displaystyle =\displaystyle 8Substituting x=0
\displaystyle 3x\displaystyle =\displaystyle 8Simplifying the 0y term
\displaystyle x\displaystyle =\displaystyle \dfrac{8}{3}Isolating x

Therefore, the zero of the function occurs at \left(\dfrac{8}{3},\,0\right).

Example 3

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

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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Idea summary

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

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

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

Outcomes

A.F.1

The student will investigate, analyze, and compare linear functions algebraically and graphically, and model linear relationships.

A.F.1a

Determine and identify the domain, range, zeros, slope, and intercepts of a linear function, presented algebraically or graphically, including the interpretation of these characteristics in contextual situations.

A.F.1c

Write equivalent algebraic forms of linear functions, including slope-intercept form, standard form, and point-slope form, and analyze and interpret the information revealed by each form.

A.F.1di

Write the equation of a linear function to model a linear relationship between two quantities, including those that can represent contextual situations. Writing the equation of a linear function will include the following situations: i) given the graph of a line;

A.F.1dii

Write the equation of a linear function to model a linear relationship between two quantities, including those that can represent contextual situations. Writing the equation of a linear function will include the following situations: ii) given two points on the line whose coordinates are integers;

A.F.1diii

Write the equation of a linear function to model a linear relationship between two quantities, including those that can represent contextual situations. Writing the equation of a linear function will include the following situations: iii) given the slope and a point on the line whose coordinates are integers;

A.F.1e

Write the equation of a line parallel or perpendicular to a given line through a given point.

A.F.1f

Graph a linear function in two variables, with and without the use of technology, including those that can represent contextual situations.

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