Mathematics - Point Question with Solution | TestHub

Let $\mathrm{P}(0,7,10),\mathrm{Q}(-1,6,6)$ and $\mathrm{R}(-4,9,6)$ be the vertices of a triangle Here, $\mathrm{PQ}=\sqrt{1+1+16}=3\sqrt{2}$ $\begin{array}{l}\mathrm{QR}=\sqrt{9+9+0}=3\sqrt{2}\\ \mathrm{PR}=\sqrt{16+4+16}=6\\ \text{ Now, }{\mathrm{PQ}}^2+{\mathrm{QR}}^2=(3\sqrt{2}{)}^2+(3\sqrt{2}{)}^2=36=(\mathrm{PR}{)}^2\end{array}$ Therefore, $△\mathrm{PQR}$ is a right angled triangle at $\mathrm{Q}$. Also, $\mathrm{OQ}=\mathrm{QR}$. Hence, $△\mathrm{PQR}$ is a right angled isosceles triangle.