Slope Re-Test Review
Please do your work on a SEPARATE piece of paper!!
Things you’ll need to KNOW!
 Point-Slope Form, Slope-Intercept Form, Standard Form
 Slope Formula!!
 What is an “intercept” (what does that mean) and how do you find it?
 How to find the slope of a line using two points
 How to write an equation of a line in point-slope form and convert it to slope-intercept form
 How to graph a line!!!! (ALL THREE WAYS!)
 How to write an equation for a horizontal and vertical line
 The slope of all vertical and horizontal lines
Multiple Choice
Identify the choice that best completes the statement or answers the question.
Match the equation with its graph.
____
1. 7x + 2y = 14
a.
–10 –8
y
–6
–4
10
10
8
8
6
6
4
4
2
2
–2
–2
2
4
6
8
10
x
–4
–6
–4
–2
–2
–4
–6
–6
–8
–8
–10
–10
10
10
8
8
6
6
4
4
2
2
–2
–2
2
4
6
8
10
x
2
4
6
8
10
x
2
4
6
8
10
x
y
d.
y
–6
–10 –8
–4
b.
–10 –8
y
c.
–10 –8
–6
–4
–2
–2
–4
–4
–6
–6
–8
–8
–10
–10
Find the slope of the line using the formula for slope.
2.
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
Find the slope of the line that passes through the pair of points.
3. (2, 8), (9, –3)
4. Use the slope and y-intercept to graph the equation.
3
y= x–3
4
Find the x- and y-intercept of the line.
5. –8x – 3y = 96
Graph the equation.
6. y = –3
7. x = –4
4
8. y – 2 = (x + 3)
3
Are the graphs of the lines in the pair parallel? Explain.
1
x – 11
11
x – 11y = –9
9. y =
10. y = 6x + 3
–15x + 3y = –45
Write an equation for the line that is parallel to the given line and that passes through the given point.
9
11. y = x + 9; (–5, –2)
5
Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.
3
12. y =  x + 2
4
16x – 12y = 16
Write the equation of a line that is perpendicular to the given line and that passes through the given point.
13. –x + 12y = 18; (–2, –6)
pe Re-Test
swer Section
ULTIPLE CHOICE
1. ANS:
OBJ:
NAT:
TOP:
KEY:
B
PTS: 1
DIF: L2
REF: 6-4 Standard Form
6-4.1 Graphing Equations Using Intercepts
NAEP 2005 A1h | ADP J.4.1 | ADP J.4.2 | ADP K.10.2
STA: CO 9.2.4.a | CO 9.2.1.a | CO 9.2.5.a
6-4 Example 2
graphing | x-intercept | y-intercept | standard form of a linear equation
ORT ANSWER
2. ANS:
3
PTS:
OBJ:
NAT:
STA:
KEY:
3. ANS:
11

7
1
DIF: L2
REF: 6-1 Rate of Change and Slope
6-1.2 Finding Slope
NAEP 2005 A2a | NAEP 2005 A2b | ADP J.4.1 | ADP K.10.1
CO 9.5.1.b | CO 9.2.1.a | CO 9.2.2.c
TOP: 6-1 Example 3
graphing | finding slope using a graph | slope
PTS:
OBJ:
NAT:
STA:
KEY:
4. ANS:
1
DIF: L2
REF: 6-1 Rate of Change and Slope
6-1.2 Finding Slope
NAEP 2005 A2a | NAEP 2005 A2b | ADP J.4.1 | ADP K.10.1
CO 9.5.1.b | CO 9.2.1.a | CO 9.2.2.c
TOP: 6-1 Example 4
finding slope using points | slope
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
PTS: 1
DIF: L2
REF: 6-2 Slope-Intercept Form
OBJ: 6-2.2 Graphing Linear Equations
NAT: NAEP 2005 A1h | ADP J.4.1 | ADP J.4.2 | ADP K.10.2
STA: CO 9.2.1.a | CO 9.2.4.a | CO 9.2.5.a
TOP: 6-2 Example 4
KEY: linear equation | graphing equations | slope | y-intercept
5. ANS:
x-intercept is –12; y-intercept is –32.
PTS:
OBJ:
NAT:
TOP:
KEY:
6. ANS:
1
DIF: L2
REF: 6-4 Standard Form
6-4.1 Graphing Equations Using Intercepts
NAEP 2005 A1h | ADP J.4.1 | ADP J.4.2 | ADP K.10.2
STA: CO 9.2.4.a | CO 9.2.1.a | CO 9.2.5.a
6-4 Example 1
x-intercept | y-intercept | standard form of a linear equation
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
PTS:
OBJ:
NAT:
TOP:
7. ANS:
1
DIF: L2
REF: 6-4 Standard Form
6-4.1 Graphing Equations Using Intercepts
NAEP 2005 A1h | ADP J.4.1 | ADP J.4.2 | ADP K.10.2
STA: CO 9.2.4.a | CO 9.2.1.a | CO 9.2.5.a
6-4 Example 3
KEY: graphing | horizontal and vertical lines
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
PTS:
OBJ:
NAT:
TOP:
8. ANS:
1
DIF: L2
REF: 6-4 Standard Form
6-4.1 Graphing Equations Using Intercepts
NAEP 2005 A1h | ADP J.4.1 | ADP J.4.2 | ADP K.10.2
STA: CO 9.2.4.a | CO 9.2.1.a | CO 9.2.5.a
6-4 Example 3
KEY: graphing | horizontal and vertical lines
y
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
PTS: 1
DIF: L2
REF: 6-5 Point-Slope Form and Writing Linear Equations
OBJ: 6-5.1 Using Point-Slope Form
NAT: NAEP 2005 A1h | NAEP 2005 A1i | NAEP 2005 A2a | NAEP 2005 A2b | NAEP 2005 A3a | ADP J.4.1 | ADP
J.4.2 | ADP K.10.1 | ADP K.10.2
STA:
CO 9.2.1.a | CO 9.2.5.a
TOP: 6-5 Example 1
KEY: point-slope form | graphing | linear equation
9. ANS:
Yes, since the slopes are the same and the y-intercepts are different.
PTS: 1
DIF: L2
REF: 6-6 Parallel and Perpendicular Lines
OBJ: 6-6.1 Parallel Lines
NAT: NAEP 2005 G3g | NAEP 2005 A2e | ADP K.2.1 | ADP K.2.2 | ADP K.10.1 | ADP K.10.2
STA: CO 9.2.1.a | CO 9.2.4.a
TOP: 6-6 Example 1
KEY: parallel lines | slope
10. ANS:
No, since the slopes are different.
PTS: 1
DIF: L2
REF: 6-6 Parallel and Perpendicular Lines
OBJ: 6-6.1 Parallel Lines
NAT: NAEP 2005 G3g | NAEP 2005 A2e | ADP K.2.1 | ADP K.2.2 | ADP K.10.1 | ADP K.10.2
STA: CO 9.2.1.a | CO 9.2.4.a
TOP: 6-6 Example 1
KEY: parallel lines | slope
11. ANS:
9
y= x+7
5
PTS: 1
DIF: L2
REF: 6-6 Parallel and Perpendicular Lines
OBJ: 6-6.1 Parallel Lines
NAT: NAEP 2005 G3g | NAEP 2005 A2e | ADP K.2.1 | ADP K.2.2 | ADP K.10.1 | ADP K.10.2
STA: CO 9.2.1.a | CO 9.2.4.a
TOP: 6-6 Example 2
KEY: parallel lines | linear equation
12. ANS:
perpendicular
PTS: 1
DIF: L3
REF: 6-6 Parallel and Perpendicular Lines
OBJ: 6-6.2 Perpendicular Lines
NAT: NAEP 2005 G3g | NAEP 2005 A2e | ADP K.2.1 | ADP K.2.2 | ADP K.10.1 | ADP K.10.2
STA: CO 9.2.1.a | CO 9.2.4.a
TOP: 6-6 Example 3
KEY: perpendicular lines | parallel lines
13. ANS:
y = 12x – 30
PTS:
OBJ:
NAT:
STA:
KEY:
1
DIF: L2
REF: 6-6 Parallel and Perpendicular Lines
6-6.2 Perpendicular Lines
NAEP 2005 G3g | NAEP 2005 A2e | ADP K.2.1 | ADP K.2.2 | ADP K.10.1 | ADP K.10.2
CO 9.2.1.a | CO 9.2.4.a
TOP: 6-6 Example 3
perpendicular lines | linear equation