Prove that the points $(7, 10), (-2, 5)$ and $(3, -4)$ are the vertices of an isosceles right triangle.

Given:

Given points are $(7, 10), (-2, 5)$ and $(3, -4)$.

To do:

We have to prove that the points $(7, 10), (-2, 5)$ and $(3, -4)$ are the vertices of an isosceles right triangle.

Solution:

 Vertices of a \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(7,10), \mathrm{B}(-2,5) \) and \( \mathrm{C}(3,-4) \).

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( \mathrm{AB}=\sqrt{(7+2)^{2}+(10-5)^{2}} \)

\( =\sqrt{(9)^{2}+(5)^{2}} \)

\( =\sqrt{81+25} \)

\( =\sqrt{106} \)

Similarly,

\( \mathrm{BC}=\sqrt{(-2-3)^{2}+(5+4)^{2}} \)

\( =\sqrt{(-5)^{2}+(9)^{2}} \)

\( =\sqrt{25+81} \)

\( =\sqrt{106} \)

\( \mathrm{CA}=\sqrt{(7-3)^{2}+(10+4)^{2}} \)

\( =\sqrt{(4)^{2}+(14)^{2}} \)

\( =\sqrt{16+196} \)

\( =\sqrt{212} \)

Here,

\( \mathrm{AB}=\mathrm{BC} \) and \( \mathrm{CA} \) is the longest side.

\( \mathrm{AB}^{2}+\mathrm{BC}^{2}=(\sqrt{106})^{2}+(\sqrt{106})^{2} \)

\( =106+106=212 \)

\( \mathrm{CA}^{2}=(\sqrt{212})^{2}=212 \)

\( \Rightarrow \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)

Therefore, by Pythagoras theorem, $(7, 10), (-2, 5)$ and $(3, -4)$ are the vertices of an isosceles right triangle.

Hence proved.

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Updated on: 10-Oct-2022

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