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Prove that the points $(0, 0), (5, 5)$ and $(-5, 5)$ are the vertices of a right isosceles triangle.
Given:
Given points are $(0, 0), (5, 5)$ and $(-5, 5)$.
To do:
We have to prove that the points $(0, 0), (5, 5)$ and $(-5, 5)$ are the vertices of a right isosceles triangle.
Solution:
Let the vertices of the \( \Delta \mathrm{ABC} \) be \( \mathrm{A}(0,0), \mathrm{B}(5,5) \) and \( \mathrm{C}(-5,5) \).
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,
\( A B=\sqrt{(5-0)^{2}+(5-0)^{2}} \)
\( =\sqrt{(5)^{2}+(5)^{2}} \)
\( =\sqrt{25+25} \)
\( =\sqrt{50} \)
\( B C=\sqrt{(-5-5)^{2}+(5-5)^{2}} \)
\( =\sqrt{(-10)^{2}+(0)^{2}} \)
\( =\sqrt{100} \)
\( =10 \)
\( CA=\sqrt{(0+5)^{2}+(0-5)^{2}} \)
\( =\sqrt{(5)^{2}+(-5)^{2}} \)
\( =\sqrt{25+25} \)
\( =\sqrt{50} \)
Here,
\( \mathrm{AB}=\mathrm{CA} \) and \( \mathrm{BC} \) is the longest side.
\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{50})^{2}+(\sqrt{50})^{2} \)
\( =50+50=100 \)
\( \mathrm{BC}^{2}=(10)^{2}=100 \)
\( \therefore \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)
Therefore, \( \Delta \mathrm{ABC} \) is an right isosceles triangle.
Hence proved.
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