Question

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are △xyz and △hij congruent?

Explanation:

Step1: Calculate the side lengths of \(\triangle XYZ\)

• \(XZ = 10\) (from \(x = 0\) to \(x=- 10\) on the \(x -\)axis)
• Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(XY\): \(X(1,0)\), \(Y(-4,6)\)

\(XY=\sqrt{(-4 - 1)^2+(6 - 0)^2}=\sqrt{25 + 36}=\sqrt{61}\)

• For \(YZ\): \(Y(-4,6)\), \(Z(-10,0)\)

\(YZ=\sqrt{(-10+4)^2+(0 - 6)^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\)

Step2: Calculate the side lengths of \(\triangle HIJ\)

• \(HJ=10\) (horizontal length from \(x = 0\) to \(x = 10\))
• For \(HI\): \(H(0,9)\), \(I(4,5)\)

\(HI=\sqrt{(4 - 0)^2+(5 - 9)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}\)

• For \(IJ\): \(I(4,5)\), \(J(10,9)\)

\(IJ=\sqrt{(10 - 4)^2+(9 - 5)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\)

Step3: Compare side - lengths

Since \(XY
eq HI\), \(YZ
eq IJ\) (and we can also check angles using slope - formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), slope of \(XY\) is \(\frac{6-0}{-4 - 1}=-\frac{6}{5}\), slope of \(HI\) is \(\frac{5 - 9}{4-0}=-1\), so angles are not equal)

Answer:

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