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1. which point is collinear to points a and d?
image of a diagram with points a, b, c, d, e, f and lines
a. b
b. c
c. e
d. f
2. if the list below represents four non - coplanar points from the diagram, which two points could complete the list?
image of a rectangular prism with vertices p, q, r, s, t, u, v, w p, t, __, __
a. w, s
b. v, r
c. q, r
d. q, u
3. given l(-4, 3) and m(-7, -2), what is the length of $\overline{lm}$?
a. $\sqrt{34}$
b. $\sqrt{8}$
c. $\sqrt{74}$
d. $\sqrt{122}$
4. line segment ef has endpoints e(5, -1) and f(-3, 8). what are the coordinates of the midpoint of $\overline{ef}$?
a. (4, -4.5)
b. (4, 3.5)
c. (1, -4.5)
d. (1, 3.5)
5. a segment has a midpoint at (2, -4) and an endpoint at (5, -9). what are the coordinates of the other endpoint?
a. (3, 5)
b. (-1, 1)
c. (-1.5, 2.5)
d. (3.5, -6.5)
6. if k is the midpoint of $\overline{jl}$, jk = 9x - 1 and kl = 2x + 27, find jl.
a. 35
b. 70
c. 62
d. 48
7. find the value of x in the figure below.
image of two intersecting lines with angles (8x + 5)° and (13x - 14)°
a. 7
b. 8
c. 9
d. 10
8. find m∠xyv in the figure below.
image of two intersecting lines with angles (3x + 23)° at wy and (7x - 37)° at yz
a. 65°
b. 68°
c. 112°
d. 115°

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall collinear definition
Collinear points lie on the same straight line.
## Step2: Analyze the diagram
Points A, D, and F are on the same line (the line with arrowheads through A, D, F). Points B, C, E are on other lines.
# Answer:
D. F

### Question 2
# Explanation:
## Step1: Recall non - coplanar points
Non - coplanar points do not lie on the same plane. In a rectangular prism (the diagram), points P, T, and then we need two points that are not on the same plane as P and T.
## Step2: Analyze options
- Option A: W and S are on the top and bottom faces, but P, T, W, S might be coplanar? No, wait, P is on the bottom front, T is on the top front. W is on the top back, S is on the bottom back. Wait, no, the list is four non - coplanar points. Wait, the original list is P, T, _, _. The rectangular prism has vertices P, Q, R, S (bottom face) and T, U, V, W (top face). Non - coplanar points: P (bottom front), T (top front), and then we need two points from different "layers" and not in the same plane. Wait, the correct option should be such that the four points are non - coplanar. Let's see the options:
- Option B: V and R. V is top back right, R is bottom back right. P (bottom front left), T (top front left), V (top back right), R (bottom back right) – these are non - coplanar. Wait, no, maybe I made a mistake. Wait, the key is that in a rectangular prism, points that are not on the same face or the same "diagonal" plane. The correct answer is B. V, R? Wait, no, let's re - think. The initial points are P and T. P is (bottom front), T is (top front). We need two points such that all four are non - coplanar. The correct option is B. V, R? Wait, the answer is B. V, R? Wait, no, the correct answer is B? Wait, the options: A. W, S; B. V, R; C. Q, R; D. Q, U. Q is bottom front right, R is bottom back right, U is top back right, V is top back right? No, U and V are the same? Wait, the diagram: P - Q - R - S (bottom), T - U - V - W (top). So P (bottom front left), T (top front left), V (top back right), R (bottom back right) – these four points are non - coplanar. So the answer is B. V, R.
# Answer:
B. V, R

### Question 3
# Explanation:
## Step1: Use distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
## Step2: Substitute values
For $L(-4,3)$ and $M(-7,-2)$, $x_1=-4,y_1 = 3,x_2=-7,y_2=-2$
$d=\sqrt{(-7+4)^2+(-2 - 3)^2}=\sqrt{(-3)^2+(-5)^2}=\sqrt{9 + 25}=\sqrt{34}$
# Answer:
A. $\sqrt{34}$

### Question 4
# Explanation:
## Step1: Use mid - point formula
The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$
## Step2: Substitute values
For $E(5,-1)$ and $F(-3,8)$, $x_1 = 5,y_1=-1,x_2=-3,y_2 = 8$
$x$ - coordinate of mid - point: $\frac{5+( - 3)}{2}=\frac{2}{2}=1$
$y$ - coordinate of mid - point: $\frac{-1 + 8}{2}=\frac{7}{2}=3.5$
# Answer:
D. $(1,3.5)$

### Question 5
# Explanation:
## Step1: Use mid - point formula
Let the mid - point be $M(2,-4)$ and one endpoint be $A(5,-9)$ and the other endpoint be $B(x,y)$
The mid - point formula is $M=(\frac{x_A+x_B}{2},\frac{y_A + y_B}{2})$
## Step2: Solve for $x$ and $y$
For $x$ - coordinate: $2=\frac{5 + x}{2}$ – multiply both sides by 2: $4=5 + x$ – $x=4 - 5=-1$
For $y$ - coordinate: $-4=\frac{-9 + y}{2}$ – multiply both sides by 2: $-8=-9 + y$ – $y=-8 + 9 = 1$
# Answer:
B. $(-1,1)$

### Question 6
# Explanation:
## Step1: Recall mid - point property
If $K$ is the mid - point of $\overline{JL}$, then $JK=KL$
## Step2: Solve for $x$
Given $JK = 9x-1$ and $KL=2x + 27$
Set $9x-1=2x + 27$
Subtract $2x$ from both sides: $7x-1 = 27$
Add 1 to both sides: $7x=28$
Divide by 7: $x = 4$
## Step3: Find $JK$ and $KL$
$JK=9(4)-1=36 - 1 = 35$
$KL=2(4)+27=8 + 27 = 35$
## Step4: Find $JL$
$JL=JK + KL=35+35 = 70$
# Answer:
B. 70

### Question 7
# Explanation:
## Step1: Recall vertical angles or linear pair? Wait, the two angles $(8x + 5)^{\circ}$ and $(13x-14)^{\circ}$ are vertical angles? No, they are vertical angles (opposite angles formed by intersecting lines), so they are equal.
## Step2: Solve for $x$
Set $8x + 5=13x-14$
Subtract $8x$ from both sides: $5 = 5x-14$
Add 14 to both sides: $19 = 5x$? Wait, no, that's wrong. Wait, maybe they are supplementary? Wait, no, the diagram shows two angles formed by intersecting lines, they are vertical angles, so they should be equal. Wait, $8x + 5=13x-14$
$14 + 5=13x-8x$
$19 = 5x$? No, that can't be. Wait, maybe I made a mistake. Wait, the two angles are vertical angles, so $8x + 5=13x-14$
$13x-8x=5 + 14$
$5x=19$? No, the options are 7,8,9,10. So maybe they are supplementary? Wait, no, vertical angles are equal. Wait, maybe the angles are adjacent and form a linear pair? No, the diagram shows two angles opposite each other (vertical angles). Wait, maybe the problem is that the angles are equal, so $8x + 5=13x-14$
$13x-8x=5 + 14$
$5x=19$ – no, that's not matching the options. Wait, maybe I misread the angles. Wait, the angles are $(8x + 5)^{\circ}$ and $(13x-14)^{\circ}$, maybe they are supplementary? Wait, if they are adjacent and form a linear pair, then $(8x + 5)+(13x-14)=180$
$21x-9 = 180$
$21x=189$
$x = 9$
Ah! I made a mistake earlier. They are adjacent angles forming a linear pair (supplementary), not vertical angles.
## Step2: Solve for $x$
$(8x + 5)+(13x-14)=180$
$21x-9 = 180$
$21x=180 + 9=189$
$x=\frac{189}{21}=9$
# Answer:
C. 9

### Question 8
# Explanation:
## Step1: Recall vertical angles property
The angles $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles? No, wait, they are vertical angles? Wait, no, the angles $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles? Wait, no, in the diagram, $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles? Wait, no, the lines intersect at Y. So $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles? Wait, no, they are equal? Wait, no, if they are vertical angles, they are equal. Wait, no, maybe they are supplementary? Wait, no, vertical angles are equal. Wait, let's set $3x + 23=7x-37$
## Step2: Solve for $x$
$3x+23 = 7x-37$
$23 + 37=7x-3x$
$60 = 4x$
$x = 15$
## Step3: Find $m\angle XYW$
Now, $m\angle XYW=(3x + 23)^{\circ}$. Substitute $x = 15$: $3(15)+23=45 + 23 = 68^{\circ}$? Wait, no, wait, the angle we need is $m\angle XYV$. Wait, $\angle XYV$ and $\angle WYZ$ are vertical angles? Wait, no, let's re - look. The angle $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles. So $3x+23=7x - 37$ gives $x = 15$. Then $m\angle XYV$ is equal to $(3x + 23)^{\circ}$? Wait, no, $\angle XYV$ and the angle $(7x-37)^{\circ}$ – wait, no, the angle we need is $m\angle XYV$. Wait, when $x = 15$, $7x-37=7\times15-37=105 - 37 = 68^{\circ}$? Wait, no, the options are $65^{\circ}$, $68^{\circ}$, $112^{\circ}$, $115^{\circ}$. Wait, maybe I made a mistake. Wait, the two angles $(3x + 23)^{\circ}$ and $(7x-37)^{\circ}$ are vertical angles, so they are equal. So $3x+23=7x - 37$
$4x=60$
$x = 15$
Then $m\angle XYV=(3x + 23)^{\circ}=3\times15 + 23=68^{\circ}$
# Answer:
B. $68^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall collinear definition

Step2: Analyze the diagram

Step1: Recall non - coplanar points

Step2: Analyze options

Step1: Use distance formula

Step2: Substitute values

Answer:

Question 2