Regents Exam Questions Name: ________________________
G.GPE.B.5: Parallel and Perpendicular Lines 8
www.jmap.org
1
G.GPE.B.5: Parallel and Perpendicular Lines 8
1 What is an equation of the perpendicular bisector
of the line segment shown in the diagram below?
1)
y
+
2
x
=
0
2)
y
−
2
x
=
0
3)
2
y
+
x
=
0
4)
2
y
−
x
=
0
2 The coordinates of the endpoints of
AB
are
A(0,0)
and
B(0,6)
. The equation of the perpendicular
bisector of
AB
is
1)
x
=
0
2)
x
=
3
3)
y
=
0
4)
y
=
3
3 Which equation represents the perpendicular
bisector of
AB
whose endpoints are
A(8,2)
and
B(0,6)
?
1)
y
=
2
x
−
4
2)
y = −
1
2
x + 2
3)
y = −
1
2
x + 6
4)
y
=
2
x
−
1
2
4 Line segment NY has endpoints
N(−11,5)
and
Y(5, −7)
. What is the equation of the perpendicular
bisector of
NY
?
1)
y + 1 =
4
3
(x + 3)
2)
y + 1 = −
3
4
(x + 3)
3)
y − 6 =
4
3
(x − 8)
4)
y − 6 = −
3
4
(x − 8)
5 Segment JM has endpoints
J(−5,1)
and
M(7, −9)
.
An equation of the perpendicular bisector of
JM
is
1)
y − 4 =
5
6
(x + 1)
2)
y + 4 =
5
6
(x − 1)
3)
y − 4 =
6
5
(x + 1)
4)
y + 4 =
6
5
(x − 1)
6 The endpoints of
AB
are
A(0,4)
and
B(−4,6)
.
Which equation of a line represents the
perpendicular bisector of
AB
?
1)
y = −
1
2
x + 4
2)
y
=
−
2
x
+
1
3)
y
=
2
x
+
8
4)
y
=
2
x
+
9
Regents Exam Questions Name: ________________________
G.GPE.B.5: Parallel and Perpendicular Lines 8
www.jmap.org
2
7 Triangle ABC has vertices
A(0,0)
,
B(6,8)
, and
C(8, 4)
. Which equation represents the
perpendicular bisector of
BC
?
1)
y
=
2
x
−
6
2)
y
=
−
2
x
+
4
3)
y =
1
2
x +
5
2
4)
y = −
1
2
x +
19
2
8 If
AB
is defined by the endpoints
A(4,2)
and
B(8,6)
, write an equation of the line that is the
perpendicular bisector of
AB
.
9 Write an equation of the line that is the
perpendicular bisector of the line segment having
endpoints
(3,−1)
and
(3,5)
. [The use of the grid
below is optional]
10 Write an equation of the perpendicular bisector of
the line segment whose endpoints are
(−1,1)
and
(7,−5)
. [The use of the grid below is optional]
11 Determine the distance between point
A(−1,−3)
and point
B(5,5)
. Write an equation of the
perpendicular bisector of
AB
. [The use of the
accompanying grid is optional.]
ID: A
1
G.GPE.B.5: Parallel and Perpendicular Lines 8
Answer Section
1 ANS: 4
The segment’s midpoint is the origin and slope is
−2
. The slope of a perpendicular line is
1
2
.
y =
1
2
x + 0
2y = x
2y − x = 0
REF: 081724geo
2 ANS: 4
AB
is a vertical line, so its perpendicular bisector is a horizontal line through the midpoint of
AB
, which is
(0,3)
.
REF: 011225ge
3 ANS: 1
m =
8 + 0
2
,
2 + 6
2
= (4,4)
m =
6 − 2
0 − 8
=
4
−8
= −
1
2
m
⊥
= 2
y = mx + b
4 = 2(4) + b
−4 = b
REF: 081126ge
4 ANS: 1
m =
−11 + 5
2
,
5 +−7
2
= (−3,−1)
m =
5 −−7
−11 − 5
=
12
−16
= −
3
4
m
⊥
=
4
3
REF: 061612geo
5 ANS: 4
−5 + 7
2
,
1 − 9
2
= (1,−4)
m =
1 −−9
−5 − 7
=
10
−12
= −
5
6
m
⊥
=
6
5
REF: 062220geo
6 ANS: 4
−4 + 0
2
,
6 + 4
2
→ (−2,5)
;
6 − 4
−4 − 0
=
2
−4
= −
1
2
;
m
⊥
= 2
;
y − 5 = 2(x + 2)
y = 2x + 4 + 5
y = 2x + 9
REF: 062324geo
ID: A
2
7 ANS: 3
midpoint:
6 + 8
2
,
8 + 4
2
= (7,6)
. slope:
8 − 4
6 − 8
=
4
−2
= −2; m
⊥
=
1
2
.
6 =
1
2
(7) + b
12
2
=
7
2
+ b
5
12
= b
REF: 081327ge
8 ANS:
M =
4 + 8
2
,
2 + 6
2
= (6,4)
m =
6 − 2
8 − 4
=
4
4
= 1
m
⊥
= −1
y − 4 = −(x − 6)
REF: 081536ge
9 ANS:
M =
3 + 3
2
,
−1 + 5
2
= (3,2)
.
y = 2
.
REF: 011334ge
10 ANS:
y =
4
3
x − 6
.
M
x
=
−1 + 7
2
= 3
M
y
=
1 + (−5)
2
= −2
m =
1 − (−5)
−1 − 7
= −
3
4
The perpendicular bisector goes through
(3,−2)
and has a slope of
4
3
.
y − y
M
= m(x − x
M
)
y + 2 =
4
3
(x − 3)
.
REF: 080935ge
ID: A
3
11 ANS:
10,
y − 1 = −
3
4
(x − 2)
. . To find the equation of the perpendicular bisector, calculate
midpoint and slope.
. The perpendicular bisector of goes through (2,1) and has a
slope of
. .
REF: 080235a
