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The points $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ form
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Verified Answer
The correct answer is:
a right angled isosceles triangle
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, $\triangle \mathrm{PQR}$ is a right angled triangle at $\mathrm{Q}$. Also, $\mathrm{OQ}=\mathrm{QR}$. Hence, $\triangle \mathrm{PQR}$ is a right angled isosceles triangle.
$\begin{array}{l}
\mathrm{QR}=\sqrt{9+9+0}=3 \sqrt{2} \\
\mathrm{PR}=\sqrt{16+4+16}=6 \\
\text { Now, } \mathrm{PQ}^{2}+\mathrm{QR}^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=36=(\mathrm{PR})^{2}
\end{array}$
Therefore, $\triangle \mathrm{PQR}$ is a right angled triangle at $\mathrm{Q}$. Also, $\mathrm{OQ}=\mathrm{QR}$. Hence, $\triangle \mathrm{PQR}$ is a right angled isosceles triangle.
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