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Prove that $(2, -2), (-2, 1)$ and $(5, 2)$ are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.
Given:
Given points are $(2, -2), (-2, 1)$ and $(5, 2)$.
To do:
We have to prove that the points $(2, -2), (-2, 1)$ and $(5, 2)$ are the vertices of a right angled triangle and find the area of the triangle and the length of the hypotenuse.
Solution:
Let the vertices of the triangle be \( \mathrm{A}(2,-2), \mathrm{B} \) (-2,1) and \( C(5,2) \).
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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(-2-2)^{2}+(1+2)^{2}}=(-4)^{2}+(3)^{2} \)
\( =\sqrt{16+9} \)
\( =\sqrt{25}=5 \)
\( \mathrm{BC}=\sqrt{(5+2)^{2}+(2-1)^{2}} \)
\( =\sqrt{(7)^{2}+(1)^{2}} \)
\( =\sqrt{49+1}=\sqrt{50} \)
\( \mathrm{CA}=\sqrt{(3+1)^{2}+(0-3)^{2}} \)
\( =\sqrt{(4)^{2}+(-3)^{2}} \)
\( =\sqrt{16+9} \)
\( =\sqrt{25}=5 \)
Here,
\( \mathrm{AB}=\mathrm{CA} \) and \( \mathrm{BC} \) is the longest side.
\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=5^{2}+5^{2} \)
\( =25+25=50 \)
\( \mathrm{BC}^{2}=(\sqrt{50})^2=50 \)
\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)
Therefore, \( \Delta \mathrm{ABC} \) is a right angled triangle.
Area of a triangle $=\frac{1}{2} \times $ Base $\times$ Altitude
\( =\frac{1}{2} \times 5 \times 5 \)
\( =\frac{25}{2}=12.5 \)sq. units
Length of the hypotenuse \( =\mathrm{BC}=\sqrt{50} \) units
\( =\sqrt{25+2}=5 \sqrt{2} \) units
The area of the triangle is $12.5$ sq. units and the length of the hypotenuse is $5\sqrt{2}$ units.
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