Show that the points

Solution:

To prove: Points A, B, C form isosceles triangle.

Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$

Here,

$\left(x_{1}, y_{1}, z_{1}\right)=(4,6,-3)$

$\left(x_{2}, y_{2}, z_{2}\right)=(0,2,3)$

$\left(x_{3}, y_{3}, z_{3}\right)=(-4,-4,-1)$

Length $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

$=\sqrt{(0-4)^{2}+(2-6)^{2}+(3-(-3))^{2}}$

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

$=\sqrt{16+16+36}$

$=\sqrt{68}=2 \sqrt{17}$

Length $B C=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

$=\sqrt{(-4-0)^{2}+(-4-2)^{2}+(-1-3)^{2}}$

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

$=\sqrt{16+36+16}$

$=\sqrt{68}=2 \sqrt{17}$

Length $A C=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}}$

$=\sqrt{(-4-4)^{2}+(-4-6)^{2}+(-1-(-5))^{2}}$

$=\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}$

$=\sqrt{64+100+4}$

$=\sqrt{168}$

Here, AB = BC

∴ vertices A, B, C forms an isosceles triangle