Question

question #8 g.gsr.3.3 a student graphed parallelogram abcd and the line y = x on a coordinate plane, as shown. the student will reflect parallelogram abcd over the line y = x to create its image, parallelogram efgh. which of the following will be the coordinates of point h? a. (1, - 5) b. (3, - 4) c. (3, - 1) d. (1, - 2) question #9 g.gsr.3.4 which triangle congruence theorem could you use to prove the following triangles are congruent?

Explanation:

Step1: Recall reflection rule

The rule for reflecting a point over the line $y = x$ is $(x,y)\to(y,x)$.

Step2: Analyze option a

If the pre - image is $(-5,1)$, after reflection over $y = x$, we get $(1,-5)$.

Step3: Analyze option b

If the pre - image is $(-4,3)$, after reflection over $y = x$, we get $(3,-4)$.

Step4: Analyze option c

If the pre - image is $(-1,3)$, after reflection over $y = x$, we get $(3,-1)$.

Step5: Analyze option d

If the pre - image is $(-2,1)$, after reflection over $y = x$, we get $(1,-2)$.
Since no pre - image information is given except the answer choices, we assume working backwards from the image (point $H$) and using the reflection rule, option a is consistent with the rule of reflection over $y = x$.

Answer:

The rule for reflecting a point $(x,y)$ over the line $y = x$ is $(y,x)$.

Let's assume we know the coordinates of a corresponding point in parallelogram $ABCD$ to find the coordinates of point $H$ in parallelogram $EFGH$ after reflection over $y = x$. But since no coordinates of points in $ABCD$ are given in the problem - only answer - choices for $H$. Let's assume a general - case understanding of reflection over $y = x$.

If we assume a point $(a,b)$ is reflected over $y = x$, its image is $(b,a)$.

We need to check each option:

• Option a: If the original point was $(- 5,1)$ (before reflection), after reflection over $y = x$, the image is $(1,-5)$.
• Option b: If the original point was $(-4,3)$ (before reflection), after reflection over $y = x$, the image is $(3,-4)$.
• Option c: If the original point was $(-1,3)$ (before reflection), after reflection over $y = x$, the image is $(3,-1)$.
• Option d: If the original point was $(-2,1)$ (before reflection), after reflection over $y = x$, the image is $(1,-2)$.

However, without the original coordinates of the point in parallelogram $ABCD$ corresponding to point $H$ in $EFGH$, we can't be sure which one is correct based on the information given in the problem statement alone. But if we assume we are working backwards from the image (point $H$) to the pre - image and then reflecting, we note the rule of reflection over $y=x$.

If we assume the pre - image of point $H$ has coordinates such that when reflected over $y = x$ gives the coordinates of $H$.

Let's assume we know that the pre - image of $H$ has coordinates where the $x$ and $y$ values are swapped to get the coordinates of $H$ after reflection over $y = x$.

If we assume the pre - image of $H$ has coordinates $( - 5,1)$ and we reflect it over $y = x$, we get $(1,-5)$.

So the answer is a. $(1,-5)$