Question

does each of the rigid motions below result in △abc? select yes or no. suppose a is the line with equation x = 6, b is the line with equation x = 3, and c is the line with equation x = - 2. t_(0,10)(△abc) t_(10,0)(△abc) (r_y - axis ∘ r_a)(△abc) (r_b ∘ r_c)(△abc)

Explanation:

Step1: Analyze translation $T_{(0,10)}$

A translation $T_{(0,10)}$ moves $\triangle ABC$ 10 units up. The resulting triangle will not be $\triangle A'B'C'$. So for $T_{(0,10)}(\triangle ABC)$ the answer is No.

Step2: Analyze translation $T_{(10,0)}$

A translation $T_{(10,0)}$ moves $\triangle ABC$ 10 units to the right. The resulting triangle will not be $\triangle A'B'C'$. So for $T_{(10,0)}(\triangle ABC)$ the answer is No.

Step3: Analyze composition of reflections $(R_{y - axis}\circ R_{a})$

First, reflection $R_{a}$ (where $a:x = 6$) and then $R_{y - axis}$. The combined transformation will not result in $\triangle A'B'C'$. So for $(R_{y - axis}\circ R_{a})(\triangle ABC)$ the answer is No.

Step4: Analyze composition of reflections $(R_{b}\circ R_{c})$

The line $c:x=-2$ and $b:x = 3$. The composition of reflections $R_{b}\circ R_{c}$ is equivalent to a translation. The resulting triangle will not be $\triangle A'B'C'$. So for $(R_{b}\circ R_{c})(\triangle ABC)$ the answer is No.

Answer:

$T_{(0,10)}(\triangle ABC)$: No
$T_{(10,0)}(\triangle ABC)$: No
$(R_{y - axis}\circ R_{a})(\triangle ABC)$: No
$(R_{b}\circ R_{c})(\triangle ABC)$: No