Name: ________________________ Class: ___________________ Date: __________
ID: A
Module 4: Topic A & B Review for Exam
Slopes of Lines
Slope formula: m =
∆y
rise y 2 − y 1
=
=
run x 2 − x 1
∆x
Parallel lines have same slopes: mÄ = m
Perpendicular lines have negative reciprocal slopes: (m⊥ )(m) = −1
Dilation from origin preserves slope but changes y-intercept for all lines except for lines passing
through the origin.
____
1. Which of the following transformations on the function y = 4x − 6, would produce a line parallel to
the given line?
a.
b.
____
Rotation of 180o about the origin
Translation up 7 units
c.
d.
Reflection in the x-axis
Reflection in the line y = −x
2. Which of the following linear functions would remain unchanged under a dilation of 4 about the
origin?
a.
b.
y = −3x
y = 4x − 1
c.
d.
y = 3x − 5
y = 2x + 4
3. The lines 4y − x = 5 and y + 6 = −3(x + 1) are parallel, perpendicular, or neither? Show your work.
.
4. What is the slope of the vector that describes the translation of O ′ to O?
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Name: ________________________
ID: A
Equations of Lines
Slope-intercept form of equation of a line: y = mx + b
Point-slope form of equation of a line: y − y 1 = m(x − x 1 )
5. Which equation represents a line that is parallel to the line 6y − 18 = x ?
.
6. Which equation represents a line perpendicular to y = 2x − 8 that passes through the point (6,−8)?
.
7. Two parallel roads run through a town. When the roads are graphed on the coordinate plane, one of
the roads can be represented by the equation 6x + 2y = 8. If the other road passes through the point
(5,10), what is the equation of the second road?
.
8. Find the equation of the line perpendicular to 5x − y = −35 that shares the same y-intercept.
.
9. Write the equation of the line that contains the point (−4,9) and is:
a) Parallel to the line x = 6
b) Perpendicular to the line x = −12
c) Parallel to the line y = 6x − 12
d) Perpendicular to the line y = 2x − 14
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Name: ________________________
ID: A
Coordinate Geometry Proofs
10. Is ∆RST , where R(−2,−2) , S(2,6), and T(4,−5), a right triangle? If so, which angle is the right angle?
Justify your answer. (Use of graph paper is optional.)
11. The vertices of quadrilateral JKLM have coordinates J(−3,1) , K(1,−5) , L(7,−2) , and M(3,4). Prove that
JKLM is a parallelogram. [The use of the set of axes below is optional.]
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ID: A
Module 4: Topic A & B Review for Exam
Answer Section
1. ANS: B
2. ANS: A
3. ANS:
neither.
4y = x + 5
y=
PTS: 1
PTS: 1
m2 = −3
1
5
x+
4
4
m1 =
1
4
PTS: 1
4. ANS:
7
m=−
4
PTS: 1
5. ANS:
1
m| | =
6
PTS: 1
6. ANS:
−1
y+8 =
(x − 6)
2
PTS: 1
7. ANS:
y − 10 = −3(x − 5)
−1
x−5
2
or
y=
or
y = −3x + 25
or
y = 6x + 33
1
y = − x+7
2
PTS: 1
8. ANS:
1
y = − x + 35
5
PTS: 1
9. ANS:
a. x = −4
b. y = 9
c. y − 9 = 6(x + 4)
1
d. y − 9 = − (x + 4)
2
or
PTS: 1
1
ID: A
10. ANS:
Yes; R is the right angle
Using Distance:
RS 2 = 8 2 + 4 2 = 64 + 16 = 80
RT 2 = 3 2 + 6 2 = 9 + 36 = 45
Using Slopes:
mRS = 2
mRT = −
ST 2 = 11 2 + 2 2 = 121 + 4 = 125
1
2
∴ RS ⊥ST
PTS: 1
11. ANS:
1−4
−3 1
mJM =
=
=
−3 − 3 −6 2
m= ML =
Since both opposite sides have equal slopes and are parallel, JKLM is a
4 − −2
6
3
=
=−
3−7
−4
2
mLK =
−2 − −5 3 1
= =
7−1
6 2
mKJ =
−5 − 1 −6
3
=
=−
1 − −3
4
2
parallelogram. JM =
(−3 − 3) 2 + (1 − 4) 2 =
ML = (7 − 3) 2 + (−2 − 4) 2 =
since not all sides are congruent.
45 . JM is not congruent to ML, so JKLM is not a rhombus
52
PTS: 6
REF: 061438ge
STA: G.G.69
TOP: Quadrilaterals in the Coordinate Plane
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