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Show that the points $ (1,7),(2,4) $ and $ (5,5) $ are the vertices of an isosceles right angled triangle.
Given:
Given points are $(1, 7), (2, 4)$ and $(5, 5)$.
To do:
We have to show that the points \( (1,7),(2,4) \) and \( (5,5) \) are the vertices of an isosceles right angled triangle.
Solution:
Vertices of a \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(1,7), \mathrm{B}(2,4) \) 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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(2-1)^{2}+(4-7)^{2}} \)
\( =\sqrt{(1)^{2}+(-3)^{2}} \)
\( =\sqrt{1+9} \)
\( =\sqrt{10} \)
Similarly,
\( \mathrm{BC}=\sqrt{(5-2)^{2}+(5-4)^{2}} \)
\( =\sqrt{(3)^{2}+(1)^{2}} \)
\( =\sqrt{9+1}=\sqrt{10} \)
\( \mathrm{CA}=\sqrt{(5-1)^{2}+(5-7)^{2}} \)
\( =\sqrt{(4)^{2}+(-2)^{2}} \)
\( =\sqrt{16+4} \)
\( =\sqrt{20} \)
Here,
\( \mathrm{AB}=\mathrm{BC} \) and \( \mathrm{CA} \) is the longest side.
\( \mathrm{AB}^{2}+\mathrm{BC}^{2}=(\sqrt{10})^{2}+(\sqrt{10})^{2} \)
\( =10+10=20 \)
\( \mathrm{CA}^{2}=(\sqrt{20})^{2}=20 \)
\( \therefore \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)
Therefore, \( \Delta \mathrm{ABC} \) is an isosceles right triangle.
Hence proved.
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