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**Parallel and Perpendicular Lines**

Chapter 7 Section 7 Parallel and Perpendicular Lines

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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.

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Why It’s Important Surveying Surveyors use parallel and perpendicular lines to plan construction.

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Parallel Lines Words: If two lines have the same slope, then they are parallel. Model: Symbols: II y = 2x + 2 y = 2x - 4

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**You can always check by graphing.**

Example 1 You can always check by graphing. Determine whether the graph of the equations are parallel. y = -¾x – 2 4y = -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. 4y = -3x + 12 y = -¾x + 3 Slope = -¾ y = -¾x -2 Slope intercept form Slope = -¾

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**Determine whether the graph of the equations are parallel.**

Your Turn Determine whether the graph of the equations are parallel. y = 2x 7 = 2x – y Yes

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**Determine whether the graph of the equations are parallel.**

Your Turn Determine whether the graph of the equations are parallel. y = -3x + 3 2y = 6x – 5 No

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**What is a parallelogram???**

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

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**Determine whether figure ABCD is a parallelogram.**

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. D(4,5) Solve: AB m = -1 – 1 – (-2) 3 = = -1 A(-2,2) C(7,2) Solve: BC m = -1 – = = Solve: DC m = 2 – = = -1 B(1,-1) Solve: AD m = 4 – (-2) = =

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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.

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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.

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Example 3 Write 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 Form y – 3 = -4(x - 1) Replace (x1, y1) with (1, 3) and m with -4 y – 3 = -4x + 4 Distributive Property Add 3 to both sides y = -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

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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

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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

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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

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Example 4 Determine whether the graphs of the equations are perpendicular. y = 2/3x + 1 y = -3/2x + 2 The graphs are perpendicular because the product of their slope is y = -3/2x + 2 ∙ = -1 y = 2/3x + 1

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Your Turn Determine whether the graphs of the equations are perpendicular. y = 1/5x + 2 y = 5x + 1 No

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Your Turn Determine whether the graphs of the equations are perpendicular. y = -4x + 3 4y = x -5 Yes

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Example 5 Write 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 Add 2 to both sides y = -3x – 10 The new equation y = -3x -10. Check by substituting (-4, 2) into the equation or by graphing.

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Your Turn Write 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

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Your Turn Write 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

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**Video Examples Parallel Lines 1 Parallel Lines 2 Parellal Lines 3**

Perpendicular Lines 1 Perpendicular Line 2

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