Question

justify congruence using rigid transformations
is $\triangle abc \cong \triangle xyz$? justify your answer using rigid transformations.
identify the rigid transformations that could map
$\triangle xyz$ onto $\triangle abc$.
if $x$ and $a$ are corresponding vertices, we need to
reorient $\triangle xyz$ so that $x$ is the topmost vertex of
the triangle. reflect $\triangle xyz$ in the $x$-axis.
$x(1, -2) \to x(\quad, \quad)$
$y(0, 2) \to y(\quad, \quad)$
$z(3, 4) \to z(\quad, \quad)$

Explanation:

Step1: Reflect over x-axis rule

The rule for reflecting a point $(x,y)$ over the x-axis is $(x, -y)$.

Step2: Calculate $X'$

Apply rule to $X(1,-2)$:
$X'(1, -(-2)) = (1, 2)$

Step3: Calculate $Y'$

Apply rule to $Y(0,2)$:
$Y'(0, -(2)) = (0, -2)$

Step4: Calculate $Z'$

Apply rule to $Z(3,4)$:
$Z'(3, -(4)) = (3, -4)$

Answer:

$X'(1, 2)$
$Y'(0, -2)$
$Z'(3, -4)$

Additionally, after this reflection, a translation (shift left 4 units, up 2 units for $X'$ to map to $A(-3,4)$, matching the other vertices) will map $\triangle XYZ'$ onto $\triangle ABC$, proving $\triangle ABC \cong \triangle XYZ$ since rigid transformations preserve congruence.