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2. Equations & Inequalities in Two Variables
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United States of AmericaFL
Grade 912

2.04 Parallel and perpendicular lines

Lesson
Practice
Lesson

Concept summary

Knowing that two lines are parallel or perpendicular to each other gives us information about the relationship between their slopes and graphs.

Parallel lines

Two or more lines that have the same slope and never intersect. Distinct vertical lines are parallel to each other.

x
y
Perpendicular lines

Two lines that intersect at right angles. Their slopes are opposite reciprocals. Horizontal and vertical lines are also perpendicular to each other.

x
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Opposite reciprocals

Two numbers whose product is -1.

Example:

2 and -\dfrac{1}{2}

To determine if two lines are parallel or perpendicular, we will need to find the slopes of the two lines. We often do this by rearranging to y=mx+b and comparing the values of m.

If m_1 =m_2, then the two lines are parallel.

If m_1 \cdot m_2 = -1, then the two lines are perpendicular.

If the two lines are parallel, then m_1 =m_2.

If the two lines are perpendicular, then m_1 \cdot m_2 = -1.

Worked examples

Example 1

For each of the following pairs of lines, determine whether they are parallel, perpendicular, or neither.

a

y=-3x+7 and y=3x-4

Approach

We need to identify the slopes of both lines to see if they are the same, opposite reciprocals, or neither.

Solution

For y=-3x+7, the slope is m=-3.

For y=3x-4, the slope is m=3.

The slopes are not the same. The slopes are not opposite reciprocals.

The lines y=-3x+7 and y=3x-4 are neither parallel nor perpendicular.

b

3x-5y=15 and 6x-10y=60

Approach

We need to identify the slopes of both lines to see if they are the same, opposite reciprocals, or neither.

First, we want to rearrange both lines into the form y=mx+b to identify their slopes.

Solution

For 3x-5y=15,

\displaystyle 3x-5y\displaystyle =\displaystyle 15Given equation
\displaystyle -5y\displaystyle =\displaystyle -3x+15Subtract 3x from both sides
\displaystyle y\displaystyle =\displaystyle \dfrac{-3}{-5}x+\dfrac{15}{-5}Divide both sides by -5
\displaystyle y\displaystyle =\displaystyle \dfrac{3}{5}x-3Simplify

The slope for 3x-5y=15 is m=\dfrac{3}{5}.

For 6x-10y=60,

\displaystyle 6x-10y\displaystyle =\displaystyle 60Given equation
\displaystyle -10y\displaystyle =\displaystyle -6x+60Subtract 6x from both sides
\displaystyle y\displaystyle =\displaystyle \dfrac{-6x}{-10}+\dfrac{60}{-10}Divide both sides by -10
\displaystyle y\displaystyle =\displaystyle \dfrac{3}{5}x-6Simplify

The slope for 6x-10y=60 is m=\dfrac{3}{5}.

The slopes are the same, so lines 3x-5y=15 and 6x-10y=60 are parallel.

Reflection

Notice that in standard form, Ax+By=C, the values of A and B for 3x-5y=15 and 6x-10y=60 were in the same ratio, while C was different.

c

y=-2x+3 and x-2y=10

Approach

We need to identify the slopes of both lines to see if they are the same, opposite reciprocals, or neither.

We will need to rearrange x-2y=10 to slope-intercept form.

Solution

For y=-2x+3, the slope is m=-2.

For x-2y=10,

\displaystyle x-2y\displaystyle =\displaystyle 10Given equation
\displaystyle -2y\displaystyle =\displaystyle -x+10Subtract x from both sides
\displaystyle y\displaystyle =\displaystyle \dfrac{-x}{-2}+\dfrac{10}{-2}Divide both sides by -2
\displaystyle y\displaystyle =\displaystyle \dfrac{x}{2}-5Simplify

The slope of x-2y=10 is m=\dfrac{1}{2}.

The slopes are opposite reciprocals, so lines y=-2x+3 and x-2y=10 are perpendicular.

Example 2

Find the equation of the line, L_1 that is parallel to the line y=-\dfrac{5}{7}x-7 and goes through the point \left(0,4\right). Give your answer in slope-intercept form.

Approach

The slope of parallel lines are the same, so we can to identify the slope from the line y=-\dfrac{5}{7}x-7.

The point given happens to be the y-intercept.

We can substitute the slope in for m and the y-intecept in for b in y=mx+b.

Solution

The slope of y=-\dfrac{5}{7}x-7 is m=-\dfrac{5}{7}.

The y-intercept is b=4.

The equation of the line L_1 is y=-\dfrac{5}{7}x+4.

Reflection

If we had been given a point that was not the y-intercept, then we would have needed to substitute the point in for x and y and solved for b. We could also have used point-slope form and rearranged to slope-intercept form.

Outcomes

MA.912.AR.2.2

Write a linear two-variable equation to represent the relationship between two quantities from a graph, a written description or a table of values within a mathematical or real-world context.

MA.912.AR.2.3

Write a linear two-variable equation for a line that is parallel or perpendicular to a given line and goes through a given point.

MA.912.AR.2.4

Given a table, equation or written description of a linear function, graph that function, and determine and interpret its key features.

MA.912.AR.2.5

Solve and graph mathematical and real-world problems that are modeled with linear functions. Interpret key features and determine constraints in terms of the context.

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