Question

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

Explanation:

Step1: Reverse the translation rule

To find the pre - image from the image, we reverse the given translation rule $(x,y)\to(x + 1.5,y - 3.5)$. The reverse rule is $(x,y)\to(x-1.5,y + 3.5)$. This means we subtract 1.5 from the $x$ - coordinates and add 3.5 to the $y$ - coordinates of the vertices of $\triangle A'B'C'$.

Since no specific coordinates of $\triangle A'B'C'$ are given in the text (but assuming we have them from the graph that is not shown here), we would apply the reverse - translation rule to each vertex of $\triangle A'B'C'$. For a vertex $(x_1,y_1)$ of $\triangle A'B'C'$, the corresponding vertex of $\triangle ABC$ would be $(x_1-1.5,y_1 + 3.5)$. Then we would graph the triangle with these new vertices.

Since we don't have the actual graph of $\triangle A'B'C'$ with its vertex coordinates, we can't give a specific graph as an answer. But the general method is as described above.

Answer:

To find the pre - image $\triangle ABC$, apply the rule $(x,y)\to(x - 1.5,y+3.5)$ to the vertices of $\triangle A'B'C'$ and graph the resulting triangle.