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Show that the points $ (1,1),(4,4) $ and $ (6,2) $ are the vertices of a right angled triangle.
Given:
Given points are \( (1,1),(4,4) \) and \( (6,2) \).
To do:
We have to prove that the points \( (1,1),(4,4) \) and \( (6,2) \) are the vertices of a right-angled triangle.
Solution:
Let the vertices of a \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(1,1), \mathrm{B}(4,4) \) and \( \mathrm{C}(6,2) \).
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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(4-1)^{2}+(4-1)^{2}} \)
\( =\sqrt{(3)^{2}+(3)^{2}} \)
\( =\sqrt{9+9} \)
\( =\sqrt{18}=3\sqrt2 \)
Similarly,
\( \mathrm{BC}=\sqrt{(6-4)^{2}+(2-4)^{2}} \)
\( =\sqrt{(2)^{2}+(-2)^{2}} \)
\( =\sqrt{4+4}=2\sqrt{2} \)
\( \mathrm{CA}=\sqrt{(6-1)^{2}+(2-1)^{2}} \)
\( =\sqrt{(5)^{2}+(1)^{2}} \)
\( =\sqrt{25+1} \)
\( =\sqrt{26} \)
Here,
\( \mathrm{BC} \) is the longest side.
\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{18})^{2}+(\sqrt{8})^{2} \)
\( =18+8=26 \)
\( \mathrm{BC}^{2}=(\sqrt{26})^{2}=26 \)
\( \therefore \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)
Therefore, \( \Delta \mathrm{ABC} \) is a right triangle.
Hence proved.
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