1. 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?
– 4. –
What are the slope and y-intercept of each equation?
5. y = x y = x – 8 7. y = 6x 8. y = 6x – 2 1 2 4 3 2 5 7 5 5 3 5 3
6-5

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

3. 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 =
The lines are parallel. The equations have the same slope, –2, and different y-intercepts.
6-5

4. 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 Simplify. 5 2
y = x Add 3 to each side
and simplify. 5 2
6-5

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

6. 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 – Subtract 3 from each side.
–1 = b Simplify.
The equation is y = x – 1. 1 2
6-5

7. 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

8. 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

9. 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