Question

the image of trapezoid abcd has coordinates a(-2, -5), b(-1, -2), c(2, -2), and d(3, -5). it was translated by the rule t_{-1, -3}(x, y). which diagram shows the pre - image?

Explanation:

Step1: Recall the translation rule in reverse

The translation rule $T_{- 1,-3}(x,y)$ means moving a point $(x,y)$ 1 unit to the left and 3 units down. To find the pre - image, we need to move the image points 1 unit to the right and 3 units up. The reverse rule is $T_{1,3}(x,y)$.

Step2: Find the pre - image coordinates of $A'$

For $A'(-2,-5)$, using the reverse translation rule $T_{1,3}(x,y)=(x + 1,y+3)$. So $A=(-2 + 1,-5 + 3)=(-1,-2)$.

Step3: Find the pre - image coordinates of $B'$

For $B'(-1,-2)$, $B=(-1 + 1,-2 + 3)=(0,1)$.

Step4: Find the pre - image coordinates of $C'$

For $C'(2,-2)$, $C=(2 + 1,-2 + 3)=(3,1)$.

Step5: Find the pre - image coordinates of $D'$

For $D'(3,-5)$, $D=(3 + 1,-5 + 3)=(4,-2)$.

Answer:

By plotting the pre - image points $A(-1,-2)$, $B(0,1)$, $C(3,1)$ and $D(4,-2)$, you can determine which of the given diagrams shows the pre - image. (Since no specific naming of diagrams is given in the problem description, the process to find the pre - image coordinates is provided).