Question

triangle abc was translated according to the rule (x, y) → (x + 1.5,y - 3.5) to create the image △abc shown on the coordinate plane. which graph shows the pre - image, △abc?

Explanation:

Step1: Reverse the translation rule

To find the pre - image from the image, we use the reverse rule. Given the translation rule from pre - image to image is $(x,y)\to(x + 1.5,y-3.5)$, the reverse rule from image to pre - image is $(x,y)\to(x - 1.5,y + 3.5)$.

Step2: Apply the reverse rule to vertices

For example, if we have a vertex of $\triangle A'B'C'$ say $A'(x_{A'},y_{A'})$, the coordinates of the corresponding vertex $A$ of $\triangle ABC$ will be $A(x_{A'}-1.5,y_{A'}+3.5)$. Do this for all vertices of $\triangle A'B'C'$ to get the vertices of $\triangle ABC$ and then plot them to find the pre - image graph.

Since no options are given for the graphs, we can't provide a specific final answer. But the general method to find the pre - image graph is as described above.

Answer:

Step1: Reverse the translation rule

To find the pre - image from the image, we use the reverse rule. Given the translation rule from pre - image to image is $(x,y)\to(x + 1.5,y-3.5)$, the reverse rule from image to pre - image is $(x,y)\to(x - 1.5,y + 3.5)$.

Step2: Apply the reverse rule to vertices

For example, if we have a vertex of $\triangle A'B'C'$ say $A'(x_{A'},y_{A'})$, the coordinates of the corresponding vertex $A$ of $\triangle ABC$ will be $A(x_{A'}-1.5,y_{A'}+3.5)$. Do this for all vertices of $\triangle A'B'C'$ to get the vertices of $\triangle ABC$ and then plot them to find the pre - image graph.

Since no options are given for the graphs, we can't provide a specific final answer. But the general method to find the pre - image graph is as described above.