Chapter 7 Section 7

Parallel and Perpendicular Lines

What You’ll Learn

You’ll learn to write an equation of a line that is parallel or perpendicular to the graph of a given equation and that passes through a

given point.

Why It’s Important

Surveying

Surveyors use parallel and perpendicular lines to plan construction.

Parallel Lines

Words: If two lines have the same slope, then they are parallel.Model: Symbols: II

y = 2x + 2

y = 2x - 4

Example 1

Determine whether the graph of the equations are parallel. y = -¾x – 24y = -3x + 12

First, determine the slope of the lines. Write each equation in slope intercept form.

The slopes are the same, so the lines are parallel.

y = -¾x -2 Slope intercept form

Slope = -¾

4y = -3x + 12

4y = -3x + 12 4 4

y = -¾x + 3

Slope = -¾

You can always

check by graphing.

Your Turn

Determine whether the graph of the equations are parallel.

y = 2x7 = 2x – y

Yes

Your Turn

Determine whether the graph of the equations are parallel.

y = -3x + 32y = 6x – 5

No

What is a parallelogram???

A parallelogram is a four-sided figure with two sets of parallel sides.

Example 2

Determine whether figure ABCD is a parallelogram. Explore: To be a parallelogram, AB and DC must be parallel. Also, AD and BC must be parallel.

Plan: Find the slope of each segment.

A(-2,2)

B(1,-1)

C(7,2)

D(4,5)

Solve: AB m = -1 – 2 -3 1 – (-2) 3

= = -1

Solve: BC m = -1 – 2 -3 1 - 7 -6

= = ½

Solve: DC m = 2 – 5 -3 7 - 4 3

= = -1

Solve: AD m = 5 - 2 3 4 – (-2) 6

= = ½

Example 2: Continued

AB is parallel to DC because their slopes are both -1. BC is parallel to AD because their slopes are both ½. Therefore, figure ABCD is a parallelogram.

If you know the slope to a line, you can use that information to write an equation of a line that

is parallel to it.

Example 3Write an equation in slope-intercept form of the line that is parallel to the

graph y = -4x + 8 and passes through the point (1, 3).

The slope of the given line is -4. So, the slope of the new line will also be -4. Find the new equation by using the point-slope form.

y – y1 = m(x - x1) Point Slope Formy – 3 = -4(x - 1) Replace (x1, y1) with (1, 3) and m with -4y – 3 = -4x + 4 Distributive Property + 3 + 3 Add 3 to both sidesy = -4x + 7

An equation whose graph is parallel to the graph of 4x + y = 8 and passes through the point at (1, 3) is y = -4x + 7.

Checking: Check by substituting (1, 3) into y = -4x + 7 or by graphing

Your Turn

Write an equation in slope-intercept form of the line that is parallel to the graph of each

equation and passes through the given point.

y = 6x – 4; (2, 3)

y = 6x - 9

Your Turn

Write an equation in slope-intercept form of the line that is parallel to the graph of each

equation and passes through the given point.

3x + 2y = 9; (2, 0)

y = -3/2x + 3

Perpendicular Lines

Words: If the product of the slopes of two lines is -1, then the lines are perpendicular.

Model: Below

y = -x + 3

y = x -1

Example 4Determine whether the graphs of the equations are perpendicular.

y = 2/3x + 1y = -3/2x + 2

The graphs are perpendicularbecause the product of their slope is2 -3 3 2

=∙ -1

y = 2/3x + 1

y = -3/2x + 2

Your TurnDetermine whether the graphs of the equations are perpendicular.

y = 1/5x + 2y = 5x + 1

No

Your TurnDetermine whether the graphs of the equations are perpendicular.

y = -4x + 34y = x -5

Yes

Example 5Write an equation in slope-intercept form of the line that is perpendicular

to the graph of y = ⅓x – 2 and passes through the point (-4, 2).

The slope is ⅓. A line perpendicular to the graph of y = ⅓x – 2 has slope -3. Find the new equation by using the point-slope form.

y – y1 = m(x - x1) Point Slope Form

y – 2 = -3(x – (-4)) Replace (x1, y1) with (-4, 2) and m with -3

y – 2 = -3x - 12 Distributive Property + 2 + 2 Add 2 to both sidesy = -3x – 10The new equation y = -3x -10.

Check by substituting (-4, 2) into the equation or by graphing.

Your TurnWrite an equation in slope-intercept form of the line that is perpendicular

to the graph of each equation and passes through the given point.

y = 2x + 6; (0, 0)

y = -½x

Your TurnWrite an equation in slope-intercept form of the line that is perpendicular

to the graph of each equation and passes through the given point.

2x + 3y = 2; (3, 0)

y = 3/2x – 9/2

Video Examples

• Parallel Lines 1• Parallel Lines 2• Parellal Lines 3• Perpendicular Lines 1• Perpendicular Line 2