7. a line is dilated by a scale factor of $ \frac{1}{2} $ centered at a point on the line. which statement is correct about the image of the line?
a. its slope is changed by a scale factor of $ \frac{1}{2} $.
b. its y-intercept is changed by a scale factor of $ \frac{1}{2} $.
c. its slope and y-intercept are changed by a scale factor of $ \frac{1}{2} $.
d. the image of the line and the pre-image are the same.

8. after a dilation centered at the origin, the image of $ \overline{cd} $ is $ \overline{cd} $. if the coordinates of the endpoints of these segments are $ c(6, -4) $, $ d(2, -8) $, $ c(9, -6) $, and $ d(3, -12) $, the scale factor of the dilation is
a $ \frac{1}{2} $ b. $ \frac{3}{2} $ c. 3 d. $ \frac{1}{3} $

9. the line $ -3x + 4y = 8 $ is transformed by a dilation centered at the origin. which linear equation could represent its image?
a. $ y = \frac{4}{3}x + 8 $ b. $ y = \frac{3}{4}x + 8 $
c. $ y = -\frac{3}{4}x - 8 $ d. $ y = -\frac{4}{3}x - 8 $

10. triangle $ abc $ is the image of triangle $ abc $ after a dilation with a scale factor of $ \frac{1}{2} $ and centered at point $ a $. is triangle $ abc $ congruent to triangle $ abc $? explain your answer.

11. line $ mn $ is dilated by a scale factor of 2 centered at the point $ (0, 6) $. if $ \overline{mn} $ is represented by $ y = -3x + 6 $, which equation can represent $ \overline{mn} $, the image of $ \overline{mn} $?
a. $ y = -3x + 12 $ b. $ y = -3x + 6 $
c. $ y = -6x + 12 $ d. $ y = -6x + 6 $

12. triangle $ abc $ and point $ d(1, 2) $ are graphed on the set of axes below.
image of a coordinate grid with triangle abc and point d
graph and label $ \triangle abc $, the image of $ \triangle abc $, after a dilation of scale factor 2 centered at point $ d $.

Question

7. a line is dilated by a scale factor of $ \frac{1}{2} $ centered at a point on the line. which statement is correct about the image of the line?
a. its slope is changed by a scale factor of $ \frac{1}{2} $.
b. its y-intercept is changed by a scale factor of $ \frac{1}{2} $.
c. its slope and y-intercept are changed by a scale factor of $ \frac{1}{2} $.
d. the image of the line and the pre-image are the same.

8. after a dilation centered at the origin, the image of $ \overline{cd} $ is $ \overline{cd} $. if the coordinates of the endpoints of these segments are $ c(6, -4) $, $ d(2, -8) $, $ c(9, -6) $, and $ d(3, -12) $, the scale factor of the dilation is
a $ \frac{1}{2} $  b. $ \frac{3}{2} $  c. 3  d. $ \frac{1}{3} $

9. the line $ -3x + 4y = 8 $ is transformed by a dilation centered at the origin. which linear equation could represent its image?
a. $ y = \frac{4}{3}x + 8 $  b. $ y = \frac{3}{4}x + 8 $
c. $ y = -\frac{3}{4}x - 8 $  d. $ y = -\frac{4}{3}x - 8 $

10. triangle $ abc $ is the image of triangle $ abc $ after a dilation with a scale factor of $ \frac{1}{2} $ and centered at point $ a $. is triangle $ abc $ congruent to triangle $ abc $? explain your answer.

11. line $ mn $ is dilated by a scale factor of 2 centered at the point $ (0, 6) $. if $ \overline{mn} $ is represented by $ y = -3x + 6 $, which equation can represent $ \overline{mn} $, the image of $ \overline{mn} $?
a. $ y = -3x + 12 $  b. $ y = -3x + 6 $
c. $ y = -6x + 12 $  d. $ y = -6x + 6 $

12. triangle $ abc $ and point $ d(1, 2) $ are graphed on the set of axes below.
image of a coordinate grid with triangle abc and point d
graph and label $ \triangle abc $, the image of $ \triangle abc $, after a dilation of scale factor 2 centered at point $ d $.

Accepted Answer

### Question 7
# Explanation:
## Step1: Recall dilation of a line
When a line is dilated centered at a point on the line, the line remains the same (since all points on the line are scaled from the center point on the line, so the image is the same line).
## Step2: Analyze options
- Option A: Slope of a line doesn't change with dilation (dilation is a similarity transformation, preserves slope for lines). So slope isn't changed by scale factor. Eliminate A.
- Option B: y - intercept: But since the line is centered at a point on the line, the line itself doesn't change, so y - intercept (if the center is on the line) doesn't get scaled. Eliminate B.
- Option C: Slope and y - intercept aren't changed by scale factor (as dilation centered on the line keeps the line the same). Eliminate C.
- Option D: The image and pre - image are the same line. This is correct.
# Answer: D. The image of the line and the pre - image are the same.

### Question 8
# Explanation:
## Step1: Recall scale factor formula for dilation
For a dilation centered at the origin, the scale factor $k$ is given by $k=\frac{x'}{x}=\frac{y'}{y}$, where $(x,y)$ is the original point and $(x',y')$ is the image point.
## Step2: Calculate scale factor using point C
Take point $C(6, - 4)$ and its image $C'(9,-6)$. Calculate $k$ for $x$ - coordinates: $k=\frac{9}{6}=\frac{3}{2}$? Wait, no, wait. Wait, $C(6,-4)$, $C'(9, - 6)$. Wait, $\frac{9}{6}=\frac{3}{2}$? No, wait, maybe I took the wrong point. Wait, $D(2,-8)$ and $D'(3,-12)$. Let's use $D$ and $D'$. For $x$ - coordinate: $\frac{3}{2}$, for $y$ - coordinate: $\frac{- 12}{-8}=\frac{3}{2}$. Wait, but the options are $\frac{1}{2}$, $\frac{3}{2}$, $3$, $\frac{1}{3}$. Wait, $\frac{3}{2}$ is option B? Wait, no, wait, $C(6,-4)$, $C'(9,-6)$: $\frac{9}{6}=\frac{3}{2}$, $\frac{-6}{-4}=\frac{3}{2}$. $D(2,-8)$, $D'(3,-12)$: $\frac{3}{2}$, $\frac{-12}{-8}=\frac{3}{2}$. So scale factor $k = \frac{3}{2}$? Wait, no, wait the options: A. $\frac{1}{2}$, B. $\frac{3}{2}$, C. $3$, D. $\frac{1}{3}$. So the scale factor is $\frac{3}{2}$, which is option B? Wait, no, wait, maybe I made a mistake. Wait, dilation centered at origin: $(x',y')=k(x,y)$. So $k=\frac{x'}{x}$. For $C$: $x' = 9$, $x = 6$, $k=\frac{9}{6}=\frac{3}{2}$. For $D$: $x'=3$, $x = 2$, $k=\frac{3}{2}$. So the scale factor is $\frac{3}{2}$, which is option B.
# Answer: B. $\frac{3}{2}$

### Question 9
# Explanation:
## Step1: Rewrite the original equation
First, rewrite $-3x + 4y=8$ in slope - intercept form ($y=mx + b$). Solve for $y$: $4y=3x + 8$, so $y=\frac{3}{4}x + 2$. The slope of the original line is $\frac{3}{4}$ and the y - intercept is $2$.
## Step2: Analyze dilation of a line centered at origin
When a line $y = mx + b$ is dilated centered at the origin with scale factor $k$, the image of the line (if the line passes through the origin, but this line $y=\frac{3}{4}x + 2$ does not pass through the origin. Wait, no, dilation of a line centered at the origin: for a line $ax+by + c = 0$, after dilation with scale factor $k$ centered at origin, the equation becomes $ax+by+\frac{c}{k}=0$? Wait, no, better to think about the slope. Dilation centered at origin: the slope of the line remains the same (because dilation is a similarity transformation, preserves the slope of lines). So the slope of the image line should be the same as the original line, which is $\frac{3}{4}$. Now let's check the options:
- Option A: $y=\frac{4}{3}x + 8$ (slope $\frac{4}{3}$, not $\frac{3}{4}$) Eliminate.
- Option B: $y=\frac{3}{4}x + 8$ (slope $\frac{3}{4}$, let's check y - intercept. Wait, original line $y=\frac{3}{4}x + 2$. When dilated centered at origin, the y - intercept? Wait, no, the line $y=\frac{3}{4}x + 2$ can be thought of as a set of points $(x,y)$. After dilation with scale factor $k$, the image points are $(kx,ky)$. Substitute into the original equation: $-3(kx)+4(ky)=8$? No, that's not right. Wait, maybe I made a mistake. Wait, the original equation is $-3x + 4y=8$, or $y=\frac{3}{4}x + 2$. Let's take a point on the original line, say when $x = 0$, $y = 2$ (the y - intercept). After dilation centered at origin with scale factor $k$, the image of $(0,2)$ is $(0,2k)$. Let's see the options. The slope must remain $\frac{3}{4}$. Option B has slope $\frac{3}{4}$. Let's check: original line $y=\frac{3}{4}x + 2$. If we dilate centered at origin, the equation of the image line: let's take two points on the original line. For example, $(0,2)$ and $(4,5)$ (since when $x = 4$, $y=\frac{3}{4}\times4 + 2=3 + 2 = 5$). After dilation with scale factor $k$, the image points are $(0,2k)$ and $(4k,5k)$. The slope between these two points is $\frac{5k - 2k}{4k-0}=\frac{3k}{4k}=\frac{3}{4}$, same as original slope. Now, let's see the y - intercept of the image line: when $x = 0$, $y = 2k$. But in the options, option B is $y=\frac{3}{4}x + 8$. Wait, maybe the scale factor is $4$? No, wait, maybe I messed up. Wait, original equation: $-3x + 4y=8$ or $y=\frac{3}{4}x + 2$. Let's check option B: $y=\frac{3}{4}x + 8$. The slope is $\frac{3}{4}$, same as original. The y - intercept is $8$, which is $4\times2$. Wait, maybe the dilation scale factor is $4$? No, the options are about the equation. Wait, the key is that dilation centered at origin preserves the slope of the line. So the slope of the image line must be equal to the slope of the original line, which is $\frac{3}{4}$. So option B has slope $\frac{3}{4}$, so that's the answer.
# Answer: B. $y=\frac{3}{4}x + 8$

### Question 10
# Explanation:
## Step1: Recall congruence and dilation
Congruent figures have the same size and shape (corresponding sides equal, corresponding angles equal). Dilation is a transformation that changes the size of the figure (unless scale factor is $1$).
## Step2: Analyze the dilation
The scale factor of the dilation is $\frac{1}{2}$, which is not equal to $1$. So the image triangle $A'B'C'$ and the original triangle $ABC$ will have corresponding sides in the ratio of $\frac{1}{2}$ (since scale factor $k=\frac{1}{2}$). So their side lengths are not equal (the image is smaller).
## Step3: Conclusion about congruence
Since congruent figures must have equal side lengths (and equal angles), and dilation with scale factor $\frac{1}{2}$ changes the side lengths (makes them half as long), triangle $ABC$ is not congruent to triangle $A'B'C'$.
# Answer: No, triangle $ABC$ is not congruent to triangle $A'B'C'$. Dilation with a scale factor of $\frac{1}{2}$ (not equal to $1$) changes the side lengths of the triangle, and congruent figures require equal side lengths (and equal angles). Since the side lengths of $A'B'C'$ are $\frac{1}{2}$ the length of $ABC$'s sides, they are not congruent.

### Question 11
# Explanation:
## Step1: Recall dilation of a line
When a line is dilated centered at a point $(x_0,y_0)$, the slope of the line remains the same (because dilation is a similarity transformation, preserves the slope of the line). The original line $MN$ has equation $y=-3x + 6$, so its slope $m=-3$.
## Step2: Analyze the center of dilation
The center of dilation is $(0,6)$, which is the y - intercept of the original line. Let's find a point on the original line. For example, when $x = 0$, $y = 6$ (the center). When $x = 1$, $y=-3(1)+6 = 3$. So the point $(1,3)$ is on $MN$.
## Step3: Dilate the point $(1,3)$
The scale factor is $2$, centered at $(0,6)$. The formula for dilation centered at $(x_0,y_0)$ with scale factor $k$ is $(x',y')=(x_0 + k(x - x_0),y_0 + k(y - y_0))$. For the point $(1,3)$ and center $(0,6)$, $k = 2$: $x'=0+2(1 - 0)=2$, $y'=6+2(3 - 6)=6+2(-3)=6 - 6 = 0$. So the image of $(1,3)$ is $(2,0)$.
## Step4: Find the equation of the image line
We know the slope of the image line is the same as the original line, $m=-3$, and it passes through $(0,6)$ (the center, which is on the original line, so it's also on the image line? Wait, no, the center is $(0,6)$, which is on $MN$. When we dilate a line centered at a point on the line, the image line is parallel to the original line (same slope) and passes through the center? Wait, no, the original line has slope $-3$ and passes through $(0,6)$. Wait, no, the original line is $y=-3x + 6$, which passes through $(0,6)$ (y - intercept) and has slope $-3$. When we dilate centered at $(0,6)$ with scale factor $2$, the image line should have the same slope ($-3$) because dilation preserves slope for lines. Now, let's check the options:
- Option A: $y=-3x + 12$ (slope $-3$)
- Option B: $y=-3x + 6$ (same as original, but scale factor $2$ should change the line? Wait, no, wait the center is $(0,6)$, which is on the line. Let's take another point on the original line, say $(1,3)$ (as before). After dilation, the image of $(1,3)$ is $(2,0)$ (as calculated). Now, let's find the equation of the line passing through $(0,6)$ and $(2,0)$. The slope is $\frac{0 - 6}{2 - 0}=\frac{-6}{2}=-3$, which matches. Now, using point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(0,6)$ and $m=-3$, we get $y-6=-3(x - 0)$, so $y=-3x + 6$? Wait, that's the original line. Wait, that can't be. Wait, no, the point $(1,3)$ when dilated with scale factor $2$ centered at $(0,6)$: the vector from $(0,6)$ to $(1,3)$ is $(1,-3)$. Scaling this vector by $2$ gives $(2,-6)$. So the image point is $(0 + 2,6-6)=(2,0)$, which is correct. Now, the line through $(0,6)$ and $(2,0)$ has equation $y=-3x + 6$? Wait, when $x = 2$, $y=-3(2)+6=0$, which is correct. But that's the same as the original line. Wait, but the scale factor is $2$. Wait, maybe I made a mistake in the dilation formula. The formula for dilation centered at $(x_0,y_0)$ is $(x',y')=(x_0 + k(x - x_0),y_0 + k(y - y_0))$. For the point $(0,6)$ (the center), $x'=0+2(0 - 0)=0$, $y'=6+2(6 - 6)=6$, so it maps to itself. For the point $(2,0)$ on the original line? Wait, no, the original line is $y=-3x + 6$, when $x = 2$, $y=0$, so $(2,0)$ is on the original line. Wait, this is confusing. Wait, the key is that when you dilate a line centered at a point on the line, the image line is the same as the original line? But the scale factor is $2$. Wait, no, that's only if the center is on the line. Wait, a line is made up of infinitely many points. If you dilate each point on the line centered at a point on the line, the image of each point is still on the line (because the line is a straight line through the center). So the image line is the same as the original line. But that contradicts the options? Wait, no, the original line is $y=-3x + 6$, option B is $y=-3x + 6$. But let's check the slope: the slope of the original line is $-3$, and dilation doesn't change the slope. So the image line should have the same slope. So option B has the same slope and the same y - intercept? Wait, maybe the center is on the line, so the line doesn't change. So the answer is B? Wait, no, wait the center is $(0,6)$, which is the y - intercept of the original line. So when we dilate the line centered at its y - intercept with scale factor $2$, the line remains the same? That seems odd, but mathematically, since every point on the line is scaled from the center (which is on the line), the image of each point is still on the line, so the image line is the same as the original line. So the equation is still $y=-3x + 6$, which is option B.
# Answer: B. $y=-3x + 6$

### Question 12
# Explanation:
## Step1: Recall dilation centered at a point
To dilate a triangle centered at point $D(1,2)$ with scale factor $2$, we use the formula for dilation centered at $(x_0,y_0)$ with scale factor $k$: for a point $P(x,y)$, its image $P'(x_0 + k(x - x_0),y_0 + k(y - y_0))$.
## Step2: Identify coordinates of $A$, $B$, $C$
From the graph, let's assume the coordinates: Let's find the coordinates of $A$, $B$, $C$. Let's say $A$ is at $(-1,1)$, $B$ is at $(0,5)$,

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QUESTION IMAGE

Question

Explanation:

Question 7

Step1: Recall dilation of a line

Step2: Analyze options

Step1: Recall scale factor formula for dilation

Step2: Calculate scale factor using point C

Step1: Rewrite the original equation

Step2: Analyze dilation of a line centered at origin

Answer:

Question 8