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

Step2: Apply the reverse rule to vertices

For a vertex of $\triangle A'B'C'$, say $A'(x_{A'},y_{A'})$, the coordinates of $A$ (pre - image) are $x_A=x_{A'}-1.5$ and $y_A=y_{A'}+3.5$. Do this for all vertices of $\triangle A'B'C'$.

Since the graph of $\triangle A'B'C'$ is given, we would take each vertex of $\triangle A'B'C'$ and apply the rule $(x,y)\to(x - 1.5,y+3.5)$ to get the vertices of $\triangle ABC$. Without the options of graphs, we can't give a specific final graph, but the process to find the pre - image vertices is as above.

If we assume the vertices of $\triangle A'B'C'$ are $A'(0,2)$, $B'( - 1,-1)$, $C'( - 4,-1)$:
For $A'$:
$x_A=0 - 1.5=-1.5$, $y_A=2 + 3.5 = 5.5$
For $B'$:
$x_B=-1-1.5=-2.5$, $y_B=-1 + 3.5 = 2.5$
For $C'$:
$x_C=-4-1.5=-5.5$, $y_C=-1 + 3.5 = 2.5$

We would then plot the triangle with these new vertices to get $\triangle ABC$.

Since no options are provided, we can't give a final answer in terms of choosing a graph. But the general method to find the pre - image is as described above.

Answer:

Step1: Reverse the translation rule

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

Step2: Apply the reverse rule to vertices

For a vertex of $\triangle A'B'C'$, say $A'(x_{A'},y_{A'})$, the coordinates of $A$ (pre - image) are $x_A=x_{A'}-1.5$ and $y_A=y_{A'}+3.5$. Do this for all vertices of $\triangle A'B'C'$.

Since the graph of $\triangle A'B'C'$ is given, we would take each vertex of $\triangle A'B'C'$ and apply the rule $(x,y)\to(x - 1.5,y+3.5)$ to get the vertices of $\triangle ABC$. Without the options of graphs, we can't give a specific final graph, but the process to find the pre - image vertices is as above.

If we assume the vertices of $\triangle A'B'C'$ are $A'(0,2)$, $B'( - 1,-1)$, $C'( - 4,-1)$:
For $A'$:
$x_A=0 - 1.5=-1.5$, $y_A=2 + 3.5 = 5.5$
For $B'$:
$x_B=-1-1.5=-2.5$, $y_B=-1 + 3.5 = 2.5$
For $C'$:
$x_C=-4-1.5=-5.5$, $y_C=-1 + 3.5 = 2.5$

We would then plot the triangle with these new vertices to get $\triangle ABC$.

Since no options are provided, we can't give a final answer in terms of choosing a graph. But the general method to find the pre - image is as described above.