5 8 Slopes of Parallel and Perpendicular Lines

5 -8 Slopes of Parallel and Perpendicular Lines Lesson Objectives: I will be able to … • Identify and graph parallel and perpendicular lines • Write equations to describe lines parallel or perpendicular to a given line Language Objective: I will be able to … • Read, write, and listen about vocabulary, key concepts, and examples Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 29 Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 30 Example 1: Identifying Parallel Lines Identify which lines are parallel. The lines described by and both have slope . These lines are parallel. The lines described by y = x and y = x + 1 both have slope 1. These lines are parallel. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Example 2: Identifying Parallel Lines Page 31 Identify which lines are parallel. Step 1: Write each equation in slope-intercept (y = mx + b) form. 1. y = 2 x – 3 √ slope = 2. 3. 2 x + 3 y = 8 3 y = -2 x + 8 4. y + 1 = 3(x – 3) y + 1 = 3 x – 9 Step 2: Find the slope of each line. √ slope = y = 3 x – 10 √ slope = 3 Step 3: Determine whether the lines are parallel. and 2 x + 3 y = 8 are parallel because both have a slope of y=2 x– 3 . y+1=3(x– 3) Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Your Turn 2 Page 31 Identify which lines are parallel. Step 1: Write each equation in slope-intercept (y = mx + b) form. √ 1. 2. -3 x + 4 y = 32 3. y = 3 x √ slope = 4 y = 3 x + 32 √ slope = 3 4. y – 1 = 3(x + 2) y – 1 = 3 x + 6 y = 3 x + 7 √ slope = 3 Step 2: Find the slope of each line. – 3 x+4 y=32 Step 3: Determine whether the lines are parallel. y=3 x y = 3 x and y – 1 = 3(x + 2) are parallel because both have a slope of 3. (Note that and -3 x + 4 y = 32 are not parallel; they are the same line!) Holt Algebra 1 y– 1=3(x+2)

5 -8 Slopes of Parallel and Perpendicular Lines Page 32 Example 3: Geometry Application Show that JKLM is a parallelogram. Use the ordered pairs and the slope formula to find the slopes of MJ and KL. MJ is parallel to KL because they have the same slope. JK is parallel to ML because they are both horizontal. Since opposite sides are parallel, JKLM is a parallelogram. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 32 Your Turn 3 Show that the points A(0, 2), B(4, 2), C(1, – 3), D(– 3, – 3) are the vertices of a parallelogram. Use the ordered pairs and slope formula to find the slopes of AD and BC. A(0, 2) D(– 3, – 3) • • B(4, 2) C(1, – 3) AD is parallel to BC because they have the same slope. AB is parallel to DC because they are both horizontal. Since opposite sides are parallel, ABCD is a parallelogram. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 30 Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 30 Helpful Hint If you know the slope of a line, the slope of a perpendicular line will be the "opposite reciprocal. ” Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 33 Example 4: Identifying Perpendicular Lines Identify which lines are perpendicular: y = 3; x = – 2; y = 3 x; y = 3 is a horizontal line, and x = – 2 is a vertical line. These lines are perpendicular. The slope of the line described by y = 3 x is 3. The slope of the line described by is. 3 and are opposite reciprocals, so these lines are perpendicular. Holt Algebra 1 . x = – 2 y =3 x y=3

5 -8 Slopes of Parallel and Perpendicular Lines Page 33 Your Turn 4 Identify which lines are perpendicular: y = – 4; y – 6 = 5(x + 4); x = 3; y = x=3 x = 3 is a vertical line, and y = – 4 is a horizontal line. These lines are perpendicular. The slope of the line described by y – 6 = 5(x + 4) is 5. The slope of the line described by y= is 5 and y = – 4 are opposite reciprocals, so these lines are perpendicular. y – 6 = 5(x + 4) Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 34 Example 5: Geometry Application Show that ABC is a right triangle. If ABC is a right triangle, AB will be perpendicular to AC. slope of AB is perpendicular to AC because Therefore, ABC is a right triangle because it contains a right angle. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 34 Example 6: Writing Equations of Parallel and Perpendicular Lines Write an equation in slope-intercept form for the line that passes through (4, 10) and is parallel to the line described by y = 3 x + 8. Step 1 Find the slope of the line. y = 3 x + 8 The slope is 3. The parallel line also has a slope of 3. Step 2 Write the equation in point-slope form. y – y 1 = m(x – x 1) Use the pointslope form. y – 10 = 3(x – 4) Step 3 Write the equation in slope-intercept form. y – 10 = 3(x – 4) y – 10 = 3 x – 12 y = 3 x – 2 Holt Algebra 1 Distribute 3 on the right side. Add 10 to both sides.

5 -8 Slopes of Parallel and Perpendicular Lines Page 35 Example 7: Writing Equations of Parallel and Perpendicular Lines Write an equation in slope-intercept form for the line that passes through (2, – 1) and is perpendicular to the line described by y = 2 x – 5. Step 1 Find the slope of the line. Step 2 Write the equation in point-slope form. The slope is 2. y = 2 x – 5 y – y 1 = m(x – x 1) The perpendicular line has a slope of . Step 3 Write the equation in slope-intercept form. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Page 36 Your Turn 7 Write an equation in slope-intercept form for the line that passes through (– 5, 3) and is perpendicular to the line described by y = 5 x. Step 1 Find the slope of the line. The slope is 5. y = 5 x The perpendicular line has a slope of Step 2 Write the equation in point-slope form. y – y 1 = m(x – x 1) . Step 3 Write in slope-intercept form. Holt Algebra 1

5 -8 Slopes of Parallel and Perpendicular Lines Homework Assignment #19. 5 • Holt 5 -8 #9 -15 odd, 22 -24, 26, 34 -36 Chapter 5 Test in two classes! Holt Algebra 1