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 $T_{(0,10)}$

Translation $T_{(0,10)}$ moves $\triangle ABC$ 10 units up. But $\triangle A''B''C''$ is not just a vertical - up translation of $\triangle ABC$. So the result is No.

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

Translation $T_{(10,0)}$ moves $\triangle ABC$ 10 units to the right. But $\triangle A''B''C''$ is not just a horizontal - right translation of $\triangle ABC$. So the result is No.

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

First, reflection $R_{a}$ (about $x = 6$) and then reflection $R_{y - axis}$ (about $x = 0$). The combined transformation does not map $\triangle ABC$ to $\triangle A''B''C''$. So the result is No.

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

Reflection $R_{c}$ about $x=-2$ and $R_{b}$ about $x = 3$. The distance between $x=-2$ and $x = 3$ is $5$ units. The composition of two reflections over parallel lines is a translation. But this translation does not map $\triangle ABC$ to $\triangle A''B''C''$. So the result 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