Question

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are \\( \triangle xyz \\) and \\( \triangle fgh \\) congruent?

Explanation:

Step1: Calculate the side lengths of $\triangle XYZ$

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $XY$: $X(9,3)$, $Y(-1,3)$, so $XY=\sqrt{(9 - (-1))^2+(3 - 3)^2}=\sqrt{10^2}=10$.
For $YZ$: $Y(-1,3)$, $Z(4,9)$, so $YZ=\sqrt{(4 - (-1))^2+(9 - 3)^2}=\sqrt{5^2 + 6^2}=\sqrt{25+36}=\sqrt{61}$.
For $XZ$: $X(9,3)$, $Z(4,9)$, so $XZ=\sqrt{(9 - 4)^2+(3 - 9)^2}=\sqrt{5^2+(-6)^2}=\sqrt{25 + 36}=\sqrt{61}$.

Step2: Calculate the side lengths of $\triangle FGH$

For $FG$: $F(-9,-9)$, $G(-9,3)$, so $FG=\sqrt{(-9-(-9))^2+(3 - (-9))^2}=\sqrt{0^2+12^2}=12$.
For $GH$: $G(-9,3)$, $H(-3,-3)$, so $GH=\sqrt{(-3 - (-9))^2+(-3 - 3)^2}=\sqrt{6^2+(-6)^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$.
For $FH$: $F(-9,-9)$, $H(-3,-3)$, so $FH=\sqrt{(-3 - (-9))^2+(-3 - (-9))^2}=\sqrt{6^2+6^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$.

Step3: Compare side - lengths

Since $XY = 10$, $FG=12$, and $10
eq12$, the triangles are not congruent.

Answer:

No, $\triangle XYZ$ and $\triangle FGH$ are not congruent.