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Midterm Review Key
1. The slopes of perpendicular lines are negative reciprocals. So you need to find the slope of the
given line and then flip and negate it. You are given:
5x + 3y = 8 so solve for y and pick out the slope by
-5x
-5x
3y = -5x + 8
3
3
3
Y=
−5
x
3
8
+3
which has a slope =
−5
,
3
the negative reciprocal is
Therefore, the slope of the perpendicular line =
+3
5
3
5
2. Parallel lines have the same slope. Given the equation y= 2x +1. A line that is parallel to it would
have the same slope so you need to use a slope = 2 and through the point (2, -1).
When given a point and a slope you can solve for the y-intercept by using the point as x and y
and using the slope as m in y = mx +b.
So, if y = mx + b
-1 = 2(2) + b
-1 = 4 + b
-4 -4
-5 = b
Therefore, the standard form of the equation of the line parallel to y = 2x + 1 and through the
point (2, -1) is y = 2x – 5.
3. The midpoint = (
𝑥1+ 𝑥2 𝑦1+ 𝑦2
,
2
2
) so given endpoints A(7, -1) and B(-3, 3) substitute into the
midpoint formula: midpoint = (
7+ −3 −1+3
,
)
2
2
4 2
2 2
= ( , ) = (2, 1).
4. If A(-3, 6) and the midpoint = (-5, 2), you can solve for the other endpoint by setting up separate
equations to solve for the x and y parts of the coordinate for point B as follows:
𝑥1+ 𝑥2
So
Solve
So
2(-5)
-3 + x = -10
+3
𝑦1+ 𝑦2
= x part of the midpoint
2
−3+𝑥
= -5
2
−3+𝑥
for x: 2( 2 ) =
+3
Solve for y:
= y part of the midpoint
2
6+𝑦
=2
2
6+𝑦
2( 2 ) =
2(2)
6+y=4
-6
-6
X = -7
y = -2
Therefore, the midpoint = (-7, -2). So, the answer for number 4 is choice 4.
5. Distance = √(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2 = √(−1 − 3)2 + (4 − −2)2 = √(−4)2 + (6)2 =
√16 + 36 = √52 = √4 x √13 = 2√13. Final answer 2√13.
See the construction review packet for reference on any construction question
6. Choice 3 is the full illustration of the construction of an angle bisector. The other choices show
incomplete illustrations of the construction.
7. Choice 1 is the complete illustration of the construction of the perpendicular bisector.
8. The illustration shows congruent corresponding angles. Choice 4 says “When two lines are
intersected by a transversal and the corresponding angles are equal, the lines are parallel.”.
Therefore, Choice 4 is the answer (the other choices don’t even mention congruent
corresponding angles).
9. Choice 1 is the correct illustration.
10. The sum of the angles of a triangle = 180. So add the three expressions that are given and set
the sum = 180. (3x + 1) + (4x -17) + (5x – 20) = 180
3x + 4x + 5x + 1 – 17 – 20 = 180
12x – 36 = 180
+ 36 +36
12x = 216
12
12
X = 18
3x + 1 = 3(18) + 1 = 55
4x – 17 = 4(18) -17 = 55
5x – 20 = 5(18) -20 = 70
Reason Bank: If 2 angles in a triangle are congruent, then the opposite sides are congruent.
Therefore, the triangle is isosceles and choice 3 is the correct answer.
(Note: a right triangle has a 90 degree angle, a scalene triangle has 3 different degree measures
for its angles, and an equilateral triangle has all 3 angles = 60)
11. According to the diagram x = the vertex angle and 42 and y are the congruent base angles of the
given isosceles triangle. Reason bank: 2 sides of a triangle congruent, then the opposite angles
are congruent. So y = 42. If 42 + 42 + x = 180
84 + x = 180
-84
-84
Then, X = 96
So x = 96 and y = 42. Choice 4 is the correct answer.
12. The sum of the remote interior angles = The exterior angle of the triangle
So, m<A + m<B = m<ACD
(x) + (2x +15) = 5x + 5
3x + 15 = 5x + 5
-3x
-3x
15 = 2x + 5
-5
-5
10 = 2x
2 2
5=x
So, m<B = 2x + 15 = 2(5) + 15 = 25.
Choice 3 is the correct answer.
13. The sum of the two smaller sides of the triangle must be greater than the length of the third
side. So check the choices: 1) 2 + 4 is not greater than 7 so a triangle cannot be made
2) 4 + 5 is greater than 6 and all three sides have different measures,
so this is the correct choice. It is possible and it is scalene.
Choice 2
14. The sides opposite the smallest and largest angles correspond to the smallest and largest sides
of the triangle respectively. We first need to find the measure of all three angles of the triangle,
which must total 180 degrees. So, 70 + 65 + x = 180
135 + x = 180
-135
-135
X = 45
B
65
smallest side
largest side
A
45
70 C
Medium Side
So, BC < AC < AB is the correct order least to greatest
Choice 3 is correct
15. Remember “My Parents Are Aliens”
Mnemonic Device
What segments are
What is the point of
concurrent?
concurrency called?
My
Medians
Centroid
Parents
Perpendicular Bisectors
Circumcenter
Are
Angle Bisectors
Incenter
Aliens
Altitudes
Orthocenter
So , choice 4 is correct, since the perpendicular bisectors are constructed.
16. The centroid divides the parts of each median in a 2:1 ratio.
BF = 18 and is the length of the entire median so if
18
3
= 6, then PF = 6 and BP = 12.
So, choice 4 is correct.
17. The sum of the interior angles of a polygon = 180(n – 2)
(n represents the number of sides)
A hexagon has 6 sides so, 180(6 -2) = 720 total
We have 5 of the 6 angles 150 + 100 + 80 + 165 + 150 = 645
720 – 645 = 75 degrees for the remaining angle. Choice 1 is correct.
18. Each exterior angle of a regular polygon =
360
𝑛
=
360
5
= 72
So, choice 2 is correct.
19. Each interior angle of a regular polygon =
In this case,
180(𝑛−2)
𝑛
= 120 and we must solve for n.
First cross multiply so that,
Then distribute the 180
20.
21.
22.
23.
180(𝑛−2)
𝑛
180(n – 2) = 120n
180n – 360 = 120n
-120n
-120n
60n – 360 = 0
+ 360 +360
60n = 360
60
60
n = 6 sides
If you plot (5, 2) on graph paper and rotate the paper 90 degrees in a counterclockwise
direction, you will notice that the image of the point (5, 2) is now at (-2, 5). So, choice 3 is
correct. (Alternatively, you can use the rule (x, y) → (-y, x) for a 90 degree counterclockwise
rotation). Note that if you center your compass at the origin, and make a circle with the
circumference passing through (5,2), the point (-2,5) will be on the same circumference a
quarter of the way around the circle, counterclockwise.
If you plot the point (-3, 7) and reflect it across the x-axis you will see that the image of the point
will now be at (-3, -7). Choice 2 is correct. (Alternatively, you can use the rule (x, y) → (x, -y)
when reflecting the x-axis). Note: the line of reflection is the perpendicular bisector of the
segment connecting a pre-image with an image. Therefore, you can count the number of boxes
from the point (-3,7) to the x-axis and go that same distance on the other side of the x-axis.
If you plot (3, 4) and reflect it across the y-axis you will see the image of the point will now be at
(-3, 4). Choice 2 is correct. (Alternatively, you can use the rule (x, y) → (-x, y) when reflecting the
y-axis).
Remember, it is difficult to count boxes diagonally across the line y = x so we used the rule:
(x, y) → (y, x) when reflecting across the y = x line. In this case, (3, -4) → (-4, 3). Choice 4 is
correct. You can also graph the line y=x, which passes through (0,0) and has a slope of 1. If you
fold your graph paper on this line and hold your paper up to the light, you will be able to see the
location of the mirror image.
24. If you plot the point (-3, -1) and rotate the paper 180 degrees, counterclockwise, you will see
that the image of the point will now be at (3, 1). Choice 1 is correct. (Alternatively, you can use
the rule (x, y) → (-x, -y) for a rotation of 180 degrees).
25. Translations are achieved by adding the corresponding x and y parts of the translation vector to
the x and y parts of the pre-image coordinate. In this case we must first figure out what
translation vector would move P(3, 5) to 𝑃′ (6, 1).
To get from 3 to 6 we must add 3.
To get from 5 to 1 we must subtract 4.
So, the translation vector used is <3, -4>.
Now translate (-3, -5) along the vector <3, -4>. -3 + 3 = 0 and -5 + -4 = -9
So the image of the point winds up at (0, -9). Choice 1 is correct.
26. Starting at (4, 2) and first reflecting y=x (like #23) brings you to (2, 4) and then rotating that
point by 90 degrees in the counterclockwise direction (like #20) brings us to a final image point
of (-4, 2). Choice 1 is correct.
27. A dilation does not always preserve distance so that is incorrect.
A translation preserves both length and orientation so this is the correct answer. Choice 2.
(Note: Line reflections and glide reflections do not preserve orientation.)
Second 27. Careful. Look at the italicized word, not. Dilations can change the size of the figure. So Choice
3 is correct.
28. Right 3 is achieved by adding 3 to x and down 7 is achieved by subtracting 7 from y.
So, (x + 3, y – 7). Choice 1 is correct.
(Note: as a vector (x + 3, y – 7) may be written as <3, -7>)
29. A median connects a vertex of a triangle to the midpoint on the opposite side. A midpoint
divides a line segment into two congruent parts. The only choice that illustrates this point is
choice 1.
30. A perpendicular bisector forms a 90 degree angle at the midpoint of a line segment.
So, AC ≅ DC because AD is bisected at point C by definition of a segment bisector.
BC may or may not be congruent to CD . Choice 2 is correct then.
(Note: the angles in choice 3 are both 90 so they are congruent and the triangles can be proven
congruent by SAS)
31. The triangles given are already labelled so we can see that we need AG ≅ OL to give us SAS.
So, choice 2 is correct. MARK THE TRIANGLES.
32. AC  AC by reflexive. Mark this with an X. Mark ABC  CDA . AB  CD . Reason Bank:
 lines cute by a transversal  A.I.A. . The two triangles are congruent by A.A.S. Therefore,
choice 1 is correct.
33. Corresponding Parts of Congruent Triangles are Congruent. Angle C is corresponding to angle Z,
so choice 1 is wrong. Angle A is corresponding to angle X, so A  X since corresponding parts
of congruent triangles are congruent.
34. BC  BC by reflexive. Mark this with an X. AC  BC  BD  BC ; AB  CD by subtraction.
Therefore, choice 2 is correct.
35.
36.
37.
38.
39.
40. According to the givens, F, E, and D are midpoints. Therefore, AD , CE , and BF are medians.
Look back at the table in #15. The point of concurrency of the medians is the centroid.
Therefore, choice 1 is correct.
41. Again, look at the table in #15. This construction is the angle bisector construction. Therefore,
the correct answer is choice 1 because the incenter is the center of the incircle.
42.
43. -5 + 3 = -2 and 2 + -4 = -2. Therefore, the correct answer is choice 3.
44.
45.
46. Oops! Just review #15.
47. Centroid and incenter are always in. Therefore, the correct answer is choice 4.
48. The incircle passes through all three sides. Therefore, the incenter is equidistant from all three
sides. Choice 2 is correct.
49.
Reason Bank: If two parallel lines are cut by a transversal interior angles on the same side are
supplementary. Therefore,
x 2  4 x  x  30  180
x 2  5 x  150  0
( x  15)( x  10)  0
x  15; x  10
2
Let's substitute into the original expressions, BGH  x  30 and DHG  x  4 x . If x  15 ,
then BGH  15  30  15 and DHG  (15)2  4(15)  225  60  165 . If x  10 , then
BGH  10  30  40 and DHG  (10)2  4(10)  100  40  140 . The constraint is that the
angles must measure more than 0 and less than 180. The constraints are met for both values of x.
50. Mark the pair of congruent corresponding and congruent corresponding angles in your
diagrams. From our reason bank, we can prove triangles congruent by SSS, SAS, ASA, SAA, or HL.
We can eliminate SSS because we already have one pair of angles. We can also eliminate HL
because we don't have right triangles. The answer is choice 1 because it is the only choice that
will give us information we can use. We'd end up using ASA if we had the statement from choice
1. Choice 2 would give us SSA, which isn't on our reason bank. Choice 3 is completely wrong b/c
angle A doesn't correspond to angle U. Use your knowledge of rigid motions. For example, if we
translated point F to point B and then rotated triangle FLU clockwise so that FL mapped onto AB,
we could then reflect triangle FLU over line segment AB. That series of rigid motions would map
angles onto congruent angles and segments onto congruent segments. You will see than L maps
to A. U maps to T, not A. In choice 4, parallel lines will not lead to any of the pairs of angles we
studied that are formed by two parallel lines cut by a transversal.
51. The answer is choice 2. A dilation is only a rigid motion when the scale factor is 1. The rest of the
time, it is a similarity transformation, whereas, a line reflection, rotation, and a translations are
all rigid motions.
52. When we do a composite transformation using the circle between the two transformations, we
always do the listed transformations from right to left. The table below shows you the series of
images of J, R, and B under this composite transformation. Study your rules because this is not
THE midterm. It's just practice for the midterm, so we can ask you to perform any series of
transformations. Make flash cards. Google quizlet and make them online if you wish. The
answer is in the third column, so choice 3.
Point
ry  axis
T2,1
J(1,-2)
R(-3,6)
B(4,5)
(-1,-2)
(3,6)
(-4,5)
J  (1,-3)
R  (5,5)
B  (-2,4)
53. Perpendicular lines have slopes which have opposite signed reciprocals. Memorize the special
case that a horizontal line is perpendicular to a vertical line. For all other cases, you can check
your answer by getting the product of the slope of the original line and the perpendicualr line
will be -1. This is a great way to check your work. Let's get the slope of 2y=x+2 by isolating y.
1
x  1 . Using the slope2
1
1
intercept formula, y=mx+b, we can identify the slope as . The reciprocal of is 2. The
2
2
When we isolate y, we end up with the equivalent equation, y 
opposite of 2 is -2. Therefore the slope of our perpendicular line must be -2. An additional
constraint is that our line must pass through point (4,3). We can either take our point and
substitute it into choice 3 and choice 4, or we can use the point-slope formula for a linear line,
which is y  y1  m( x  x1 ) . ( x1 , y1 )  (4,3) ,and m=-2, so when we substitute, we have,
y  3  2( x  4) . Now all we have left to do is to distribute the -2, which gives us the
equivalent equation, y  3  2 x  8 and then add 3 to both sides, leaving us with
y  2 x  11 . So the correct answer is choice 3.
54. Read carefully. This question is asking us, which statement is NOT true. In similar triangles, the
ratio of the corresponding sides and corresponding perimeters is always the same. Therefore,
choice 1 and choice 4 are true. The ratio of the corresponding areas equals the square of the
ratio of the corresponding sides. Therefore, choice 3 is true. The ratio of the corresponding
angles in ALL similar triangles is always 1:1. Therefore, choice 2 is NOT true. The correct answer
is choice 2.
55. Just multiply the coordinates by the scale factor, -2. The correct answer is choice 4
56. Work backwards. The image was obtained by multiplication, so divide the image by -2 to get the
coordinates of the pre-image. The answer is choice 1.
57. Just multiply by -3, getting the image (-15,9), which is in quadrant II. The answer is choice 2.
58. The answer is choice 3 because the y-axis is the line of reflection.
59. Make three circles centered at the origin. Make the first pass through Z and Z prime, the second
pass through X and X prime, and the third pass through Y and Y prime. The answer is choice 3.
60. In order to map triangle ABC onto its image, we need a series of rigid motions. Therefore, the
answer is choice 4 because a glide reflection is the only composition.
61. First, check if the center of dilation is on your line by substituting in the coordinates of the
center into our equation. When we substitute in the coordinates of the origin, which are (0, 0),
we know that the center of dilation is not on the given line. Therefore, the dilation will be a
parallel line. Had the center been on our line, the dilation of our line would have been our line.
The scale factor is 3. The easiest way to get the image is to get the equation in slope-intercept
form and then multiply the y-intercept by the scale factor. The equation of the given line in
3
15
15
45
x  . When we get the product, (3) , we get
.
4
4
4
4
3
45
x .
Therefore, the equation of the image is y 
4
4
slope-intercept form is y 
62. If you look at the coordinates of the vertices of triangle ABC and the corresponding vertices of
triangle XYZ, you will see that the image was obtained through a reflection over the line y=x.
63. The center of dilation is on the line. There for the image is the line itself. Choice 4 is the answer.
The scale factor is irrelevant because a line is infinite and doubling something which is already
infinite doesn't change its length.
64. Since the triangle is equilateral, each side of triangle ABC is 36/3, which is 12 cm. The length of
EF is half the length of AB, which is 12. Half of 12 is 6. The correct answer is choice 1.
65. This question does not exist. Auto numbering can sometimes be a drag!
66. All you need to do here is to get the perimeter of triangle ABC and then divide by 2 or take all
three sides, divide each by 2, and then add. Whichever you prefer. I'll do the later. Half of 10 is
5. Half of 14 is 7. Half of 16 is 8. Therefore the perimeter of the triangle created by the
midsegments is 5+7+8, which is 20, our answer.
The night before review key:
5
5 3
. The reciprocal of is . The opposite
3
3 5
3
3
3
of is  . Therefore the slope of our perpendicular line must be  . An additional constraint is
5
5
5
1. See #53 if you struggle. We can identify the slope as
that our line must pass through point (15,-6).Using the point-slope formula for a linear line, which is
3
y  y1  m( x  x1 ) . ( x1 , y1 )  (15, 6) ,and m=  , so when we substitute, we have,
5
3
y  6   ( x  15) . This is the point-slope formula for the equation of our line.
5
2. Slope =
10  7 3

62 8
Length =
3.
1
 2  6 7  10  
,
   2,8 
2  
2
 2
Midpoint= 
(2  6) 2  (7  10) 2  73
x4
 x  4 y  12 
(2, 5)  
,
Make 2 equations and then solve each. 2 
and

2
2 
 2
y  12
5 
. Therefore the coordinates of B are (0,-22).
2
4.
(8 
3
3
3
3
45
3(10)
(7  8), 7  (3  7))  ( 8  (15), 7  ( 10))  ( 8  , 7 
)  ( 8  4.5, 7  3)  (3.5, 4)
10
10
10
10
10
10
5. An octagon has 8 sides
a.) (8-2)(180)=1080
6. a.) (2,-4)
e.) (-4, 2)
b.) 1080/8=135
b.) (-2,4)
c.) 360/8=45
c.) (4,2)
d.) always 360
d.) (0,9)
f.) (-6, -12)
7. Centered at the origin if they don't specify. The line doesn't pass through (0,0), therefore, we
need to get the slope and y-intercept of our given line. Keep our slope and multiply our y-intercept
by 4. Our slope is 5/7 and our y-intercept is -5. The image has the same slope of 5/7 and its yintercept is -5(4)=-20. So our equation is y 
8. a.) (x+9)(x-4)=0, so x={-9,4}
5
x  20 .
7
d.) (x+6)(x+2)=0, so x={-6,-2}
b.) (x+5)(x-8)=0, so x={-5,8}
e.) (x+5)(x-10)=0, so x={-5,10}
c.) (x+3)(x-9)=0, so x={-3,9}
f.) (x+15)(x-12)=0, so x={-15,12}