Answer
The answers are:
OPTION A
OPTION C
OPTION E
EXPLANATION
Problem Statement
The question gives us an image of a square HIJK with coordinates, H(0, 2), I(2, 0), J(0, -2), K(-2, 0) and we are told to perform the following two transformations on the image:
[tex]\begin{gathered} T_1\colon \\ y=-f(x) \\ \\ T_2\colon \\ (x,y)\to(3x,3y) \end{gathered}[/tex]
We are asked to answer questions based on these transformations
Solution
The steps we shall take to solve this question is given below:
1. Perform the first transformation T1.
2. Perform the second transformation T2.
3. Use the results from T1 and T2 to answer the questions
Step 1: Perform the first transformation T1:
We are told that the first transformation is y = - f(x). This implies that the first transformation negates all the y-values for every (x, y) coordinate point given. This has the effect of flipping the coordinate point over the x-axis.
We can perform this transformation below:
[tex]\begin{gathered} T_1\colon \\ y=-f(x) \\ \text{Given the coordinate points:} \\ H\mleft(0,2\mright),I\mleft(2,0\mright),J\mleft(0,-2\mright),K\mleft(-2,0\mright)\to H^{\prime}\mleft(0,-2\mright),I^{\prime}\mleft(2,-0\mright),J^{\prime}\mleft(0,-(-2)\mright),K\mleft(-2,-0\mright) \\ \\ \therefore T_1=H^{\prime}\mleft(0,-2\mright),I^{\prime}\mleft(2,0\mright),J^{\prime}\mleft(0,2\mright),K^{\prime}\mleft(-2,0\mright) \end{gathered}[/tex]
For better understanding, we can plot the original image HIJK along with its transformed image H'I'J'k' as follows:
The above image shows the original image (in RED) and its transformation (in GREEN). The images are over each other because the original image was flipped over the x-axis.
Step 2: Perform the second transformation T2:
The second transformation shows us that all the coordinates (x, y) from the result in step 1 are multiplied by 3.
Let us perform this transformation by hand below:
[tex]\begin{gathered} T_1=H^{\prime}(0,-2),I^{\prime}(2,0),J^{\prime}(0,2),K^{\prime}(-2,0) \\ T_2\colon \\ (x,y)\to(3x,3y) \\ H^{\prime}(0,-2)\to(3\times0,-2\times3)=(0,-6) \\ I^{\prime}(2,0)\to(3\times2,0\times3)=(6,0) \\ J^{\prime}(0,2)\to(0\times3,2\times3)=(0,6) \\ K^{\prime}(-2,0)\to(-2\times3,0\times3)=(-6,0) \\ \\ \therefore\text{The Second transformation is given as:} \\ H^{\doubleprime}(0,-6),I^{\doubleprime}(6,0),J^{\doubleprime}(0,6),K^{\doubleprime}(-6,0) \end{gathered}[/tex]
Again, we can visualize this second transformation as follows:
Step 3: Use the results from T1 and T2 to answer the questions
Now that we have performed all the relevant transformations, we can answer the questions as follows
OPTION A. H'I'J'K' ~ H"I"J"K":
This option is asking if the image, H'I'J'K' is similar to the image H"I"J"K". For two images to be similar, it means that all the angles on both images are equal. We can confirm that all the angles are 90 degrees because they are both squares.
Therefore, OPTION A is correct
OPTION B H'I'J'K' ≅ H"I"J"K"
This option is asking if the image, H'I'J'K' is exactly identical (or congruent) to the image H"I"J"K". For two images to be identical, then all the sides and angles must be equal. Image H'I'J'K' is clearly smaller than H"I"J"K", thus, implying that the two images are NOT congruent.
Therefore, Option B is incorrect.
OPTION C: y = x is a line of symmetry for HIJK and H"I"J"K"
This means that the line y=x divides the two images into two equal and symmetrical parts. We can check this by plotting the points on a graphing calculator as shown below:
The line y = x has been plotted on the images of HIJK and H"I"J" K" and is shown above. We can observe that the line y = x divides both HIJK and H"I"J" K" into two equal and symmetrical sides. Therefore, y = x is a line of symmetry for HIJK and H"I"J" K"
OPTION C IS CORRECT
Option D: H' (-2, 0):
The question is asking us to check if the point H'(-2, 0) exists as a point on the first transformation. Having performed the transformation above, we can see that H'(0, -2) was the image of the point H(0, 2).
[tex]H^{\prime}(0,-2)\ne H^{\prime}(-2,0)[/tex]
Option D is incorrect.
Option E: J"(0, 6):
This question is asking us to check if the point J"(0, 6) exists as a point on the second transformation. Having performed the transformation above, we can see that J" (0, 6) was the image to the point J'(0, 2)
[tex]J^{\doubleprime}(0,6)=J^{\doubleprime}(0,6)[/tex]
OPTION E IS CORRECT
Final Answer
The answers are:
OPTION A
OPTION C
OPTION E
