Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (For help, go to Lessons 1-6 and 6-2.) What is the reciprocal of each fraction? 1. 2. 3. – 4. – What are the slope and y-intercept of each equation? 5. y = x + 4 6. y = x – 8 7. y = 6x 8. y = 6x – 2 1 2 4 3 2 5 7 5 5 3 5 3 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 1. The reciprocal of is or 2. 2. The reciprocal of is . 3. The reciprocal of – is – . 4. The reciprocal of – is – . 5. y = x + 4 6. y = x – 8 slope: slope: y-intercept: 4 y-intercept: –8 7. y = 6x 8. y = 6x – 2 slope: 6 slope: 6 y-intercept: 0 y-intercept: – 2 1 2 4 3 5 7 Solutions 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Are the graphs of y = –2x – 1 and 4x + 2y = 6 parallel? Write 4x + 2y = 6 in slope-intercept form. Then compare with y = –2x – 1. 2y = –4x + 6 Subtract 4x from each side. y = –2x + 3 Simplify. Divide each side by 2. 2y –4x + 6 2 2 = The lines are parallel. The equations have the same slope, –2, and different y-intercepts. 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Write an equation for the line that contains (–2, 3) and is parallel to y = x – 4. 5 2 Step 1 Identify the slope of the given line. y = x – 4 slope 5 2 Step 2  Write the equation of the line through (–2, 3) using slope-intercept form. y – y1 = m(x – x1) Use point-slope form. y – 3 = (x + 2) Substitute (–2, 3) for (x1, y1) and for m. 5 2 y – 3 = x + (2) Use the Distributive Property. 5 2 y – 3 = x + 5 Simplify. 5 2 y = x + 8 Add 3 to each side and simplify. 5 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 Write an equation of the line that contains (6, 2) and is perpendicular to y = –2x + 7. Step 1 Identify the slope of the given line. y = – 2 x + 7 slope Step 2  Find the negative reciprocal of the slope. The negative reciprocal of –2 is . 1 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (continued) Step 3  Use the slope-intercept form to write an equation. y = mx + b 2 = (6) + b Substitute for m, 6 for x, and 2 for y. 1 2 1 2 2 = 3 + b Simplify. 2 – 3 = 3 + b – 3 Subtract 3 from each side. –1 = b Simplify. The equation is y = x – 1. 1 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 The line in the graph represents the street in front of a new house. The point is the front door. The sidewalk from the front door will be perpendicular to the street. Write an equation representing the sidewalk. Step 1  Find the slope m of the street. m = = = = – Points (0, 2) and (3, 0) are on the street. y2 – y1 x2 – x1 0 – 2 3 – 0 –2 3 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 (continued) Step 2 Find the negative reciprocal of the slope. The negative reciprocal of – is . So the slope of the sidewalk is . The y-intercept is –3. 2 3 The equation for the sidewalk is y = x – 3. 3 2 6-5

Parallel and Perpendicular Lines ALGEBRA 1 LESSON 6-5 3 2 1. Find the slope of a line parallel to 3x – 2y = 1. 2. Find the slope of a line perpendicular to 4x + 5y = 7. Tell whether the lines for each pair of equations are parallel, perpendicular, or neither. 3. y = 3x – 1, y = – x + 2 4. y = 2x + 5, 2x + y = –4 5. y = – x – 1, 2x + 3y = 6 6. Write an equation of the line that contains (2, 1) and is perpendicular to y = – x + 3. 5 4 1 3 perpendicular neither 2 3 parallel 1 2 y = 2x – 3 6-5