1. select all of the following transformations that preserve distance.
□ a. reflection over the y-axis
□ b. (x, y) —→ (6x, 6y)
□ c. rotation 90° clockwise about the origin
□ d. (x, y) —→ (x, −y)
□ e. dilation with scale factor 0.5 centered at the origin
□ f. (x, y) —→ (x − 8, y + 3)
2. use the law of detachment to make a conclusion.
if a point is on the perpendicular bisector of a line segment, then the point is equidistant from the segment’s endpoints.
the midpoint of $\overline{ab}$ is x.
$\overrightarrow{xy}$ is perpendicular to $\overline{ab}$.
point l is on $\overrightarrow{xy}$.
ⓐ point l is not on the perpendicular bisector of $\overline{ab}$.
ⓑ point l is equidistant to points x and y.
ⓒ point l is equidistant from points a and b.
ⓓ point l is not equidistant from points a and b.
3. in $\triangle abc$, point d is the midpoint of $\overline{ac}$. fill in the blanks to prove the isosceles triangle theorem.
proven given sas sss
$\overline{bd}$ $\overline{bc}$ $\overline{ad}$ cpctc
given: $\overline{ab} \cong \overline{bc}$
prove: $\angle a \cong \angle c$
it is $\square$ that $\overline{ab} \cong \overline{bc}$.
by the reflexive property $\overline{bd} \cong \square$. since d is the midpoint of $\overline{ac}$, $\square \cong \overline{cd}$.
so $\triangle abd \cong \triangle cbd$ by $\square$.
therefore $\angle a \cong \angle c$ by $\square$.
4. the angle bisector of $\angle abc$ is $\overrightarrow{bp}$. if $m\angle abp = 6n$, what is $m\angle abc$?
ⓐ 12
ⓑ 3n
ⓒ 12n
ⓓ 36n²

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Recall distance - preserving transformations
Distance - preserving transformations (rigid transformations) include reflections, rotations, and translations. Dilations are not distance - preserving as they scale the figure.
- **Option A**: Reflection over the $y$ - axis is a rigid transformation. It preserves the distance between points.
- **Option B**: The transformation $(x,y)\to(6x,6y)$ is a dilation with a scale factor of $6$. Dilations change the distance between points (they scale the length), so it does not preserve distance.
- **Option C**: Rotation $90^{\circ}$ clockwise about the origin is a rigid transformation. It preserves the distance between points.
- **Option D**: The transformation $(x,y)\to(x, - y)$ is a reflection over the $x$ - axis, which is a rigid transformation and preserves distance.
- **Option E**: Dilation with a scale factor of $0.5$ changes the distance between points (it scales the length down), so it does not preserve distance.
- **Option F**: The transformation $(x,y)\to(x - 8,y + 3)$ is a translation. Translations are rigid transformations and preserve distance.

## Step2: Select the correct options
Based on the above analysis, the transformations that preserve distance are A, C, D, F.
# Answer: A, C, D, F

### Problem 2
# Explanation:
## Step1: Recall the Law of Detachment
The Law of Detachment states that if we have a conditional statement $p\to q$ (where $p$ is the hypothesis and $q$ is the conclusion) and we know that $p$ is true, then we can conclude that $q$ is true.
- The conditional statement is: "If a point is on the perpendicular bisector of a line segment, then the point is equidistant from the segment’s endpoints." Here, $p$: "a point is on the perpendicular bisector of a line segment" and $q$: "the point is equidistant from the segment’s endpoints".
- We are given that the midpoint of $\overline{AB}$ is $X$ and $\overrightarrow{XY}$ is perpendicular to $\overline{AB}$, so point $L$ (on $\overrightarrow{XY}$) is on the perpendicular bisector of $\overline{AB}$ (so $p$ is true for point $L$ and segment $\overline{AB}$).
- By the Law of Detachment, we can conclude that point $L$ is equidistant from points $A$ and $B$.

## Step2: Select the correct option
Among the given options, option C states that "Point $L$ is equidistant from points $A$ and $B$".
# Answer: C

### Problem 3
# Explanation:
## Step1: Analyze the first blank
We are told that $\overline{AB}\cong\overline{BC}$. This is given to us as part of the problem statement. So the first blank should be filled with "given".
## Step2: Analyze the second blank (Reflexive Property)
The Reflexive Property states that a segment is congruent to itself. So $\overline{BD}\cong\overline{BD}$ (by Reflexive Property).
## Step3: Analyze the third blank (Midpoint property)
Since $D$ is the midpoint of $\overline{AC}$, by the definition of a midpoint, $\overline{AD}\cong\overline{CD}$.
## Step4: Analyze the fourth blank (Triangle Congruence Criterion)
We have $\overline{AB}\cong\overline{BC}$ (given), $\overline{BD}\cong\overline{BD}$ (Reflexive Property), and $\overline{AD}\cong\overline{CD}$ (midpoint). So, by the SSS (Side - Side - Side) Congruence Criterion, $\triangle ABD\cong\triangle CBD$.
## Step5: Analyze the fifth blank (CPCTC)
CPCTC (Corresponding Parts of Congruent Triangles are Congruent) states that if two triangles are congruent, then their corresponding angles are congruent. Since $\triangle ABD\cong\triangle CBD$, $\angle A\cong\angle C$ by CPCTC.
# Answer:
1. given
2. $\overline{BD}$
3. $\overline{AD}$
4. SSS
5. CPCTC

### Problem 4
# Explanation:
## Step1: Recall the angle - bisector property
An angle bisector divides an angle into two congruent angles. So, if $\overrightarrow{BP}$ is the angle bisector of $\angle ABC$, then $\angle ABP=\angle CBP$ and $\angle ABC=\angle ABP+\angle CBP$.
## Step2: Calculate $m\angle ABC$
Since $\angle ABP = 6n$ and $\angle CBP=\angle ABP = 6n$ (because $BP$ bisects $\angle ABC$), then $m\angle ABC=m\angle ABP + m\angle CBP=6n + 6n = 12n$.
# Answer: C

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Recall distance - preserving transformations

Step2: Select the correct options

Step1: Recall the Law of Detachment

Step2: Select the correct option

Step1: Analyze the first blank

Step2: Analyze the second blank (Reflexive Property)

Step3: Analyze the third blank (Midpoint property)

Step4: Analyze the fourth blank (Triangle Congruence Criterion)

Step5: Analyze the fifth blank (CPCTC)

Answer:

Problem 2