Question

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

Explanation:

Step1: Calculate side lengths using distance formula

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(\triangle GHI\):

• \(GH\): \(G(9,-2)\), \(H(9,8)\). \(d_{GH}=\sqrt{(9 - 9)^2+(8-(-2))^2}=\sqrt{0 + 100}=10\)
• \(HI\): \(H(9,8)\), \(I(2,6)\). \(d_{HI}=\sqrt{(2 - 9)^2+(6 - 8)^2}=\sqrt{49+4}=\sqrt{53}\)
• \(IG\): \(I(2,6)\), \(G(9,-2)\). \(d_{IG}=\sqrt{(9 - 2)^2+(-2 - 6)^2}=\sqrt{49 + 64}=\sqrt{113}\)

For \(\triangle XYZ\):

• \(XY\): \(X(2,0)\), \(Y(-8,0)\). \(d_{XY}=\sqrt{(-8 - 2)^2+(0 - 0)^2}=\sqrt{100}=10\)
• \(YZ\): \(Y(-8,0)\), \(Z(-6,-8)\). \(d_{YZ}=\sqrt{(-6+8)^2+(-8 - 0)^2}=\sqrt{4 + 64}=\sqrt{68}\)
• \(ZX\): \(Z(-6,-8)\), \(X(2,0)\). \(d_{ZX}=\sqrt{(2 + 6)^2+(0 + 8)^2}=\sqrt{64+64}=\sqrt{128}\)

Step2: Compare side lengths

Since \(GH = XY = 10\), but \(HI
eq YZ\) (\(\sqrt{53}
eq\sqrt{68}\)), \(IG
eq ZX\) (\(\sqrt{113}
eq\sqrt{128}\)).

Answer:

No, \(\triangle GHI\) and \(\triangle XYZ\) are not congruent.