Show that the points $ \left(1,-\frac{3}{2}\right),\left(-3,-\frac{7}{2}\right) $ and $ \left(-4,-\frac{3}{2}\right) $ are the vertices of a right angled triangle.

Given:

Given points are $ \left(1,-\frac{3}{2}\right),\left(-3,-\frac{7}{2}\right) $ and $ \left(-4,-\frac{3}{2}\right) $.

To do:

We have to prove that the points $ \left(1,-\frac{3}{2}\right),\left(-3,-\frac{7}{2}\right) $ and $ \left(-4,-\frac{3}{2}\right) $ are the vertices of a right-angled triangle.

Solution:

Let the vertices of a $ \Delta \mathrm{ABC} $ are $ \mathrm{A}(1,-\frac{3}{2}), \mathrm{B}(-3,-\frac{7}{2}) $ and $ \mathrm{C}(-4,-\frac{3}{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{(-3-1)^{2}+(-\frac{7}{2}+\frac{3}{2})^{2}} $

$ =\sqrt{(-4)^{2}+(\frac{-4}{2})^{2}} $

$ =\sqrt{16+4} $

$ =\sqrt{20} $

Similarly,

$ \mathrm{BC}=\sqrt{(-4+3)^{2}+(-\frac{3}{2}+\frac{7}{2})^{2}} $

$ =\sqrt{(-1)^{2}+(\frac{4}{2})^{2}} $

\( =\sqrt{1+4}$

$=\sqrt{5}$

$ \mathrm{CA}=\sqrt{(1+4)^{2}+(-\frac{3}{2}+\frac{3}{2})^{2}} $

$ =\sqrt{(5)^{2}+(0)^{2}} $

$ =\sqrt{25} $

$ =5 $

Here,

$ \mathrm{CA} $ is the longest side.

$ \mathrm{AB}^{2}+\mathrm{BC}^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} $

$ =20+5=25 $

$ \mathrm{CA}^{2}=(5)^{2}=25 $

$ \therefore \mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{CA}^{2} $

Therefore, $ \Delta \mathrm{ABC} $ is a right triangle.

Hence proved.

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Updated on: 10-Oct-2022

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