Name the type of triangle $PQR$ formed by the points $P(\sqrt2 , \sqrt2), Q(- \sqrt2, – \sqrt2)$ and $R (-\sqrt6 , \sqrt6 )$.

Given:

Given points are $P(\sqrt2 , \sqrt2), Q(- \sqrt2, – \sqrt2)$ and $R (-\sqrt6 , \sqrt6 )$.

To do:

We have to name the type of triangle $PQR$ formed by the given points.

Solution:

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,

\( P Q=\sqrt{(\sqrt{2}+\sqrt{2})^{2}+(\sqrt{2}+\sqrt{2})^{2}} \)
\( =\sqrt{(2 \sqrt{2})^{2}+(2 \sqrt{2})^{2}} \)

\( =\sqrt{16} \)

\( =4 \)
\( P R=\sqrt{(\sqrt{2}+\sqrt{6})^{2}+(\sqrt{2}-\sqrt{6})^{2}} \)

\( =\sqrt{2+6+2 \sqrt{12}+2+6-2 \sqrt{12}} \)

\( =\sqrt{16} \)

\( =4 \)

\( R Q=\sqrt{(-\sqrt{2}+\sqrt{6})^{2}+(-\sqrt{2}-\sqrt{6})^{2}} \)

\( =\sqrt{2+6-2 \sqrt{12}+2+6+2 \sqrt{12}} \)

\( =\sqrt{16} \)

\( =4 \)

Here,
\( P Q=P R=R Q=4, \)

Therefore, the points \( P, Q, R \) form an equilateral triangle.

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Updated on: 10-Oct-2022

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