# Show that the points

Question:

Show that the points $A(1,-1,-5), b(3,1,3)$ and $C(9,1,-3)$ are the vertices of an equilateral triangle.

Solution:

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

Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ 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,

$\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)=(1,-1,-5)$

$\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)=(3,1,3)$

$\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)=(9,1,-3)$

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

$=\sqrt{(3-1)^{2}+(1-(-1))^{2}+(3-(-5))^{2}}$

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

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

$=\sqrt{72}=6 \sqrt{2}$

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

$=\sqrt{(9-3)^{2}+(1-1)^{2}+(-3-3)^{2}}$

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

$=\sqrt{36+0+36}$

$=\sqrt{72}=6 \sqrt{2}$

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

$=\sqrt{(9-1)^{2}+(1-(-1))^{2}+(-3-(-5))^{2}}$

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

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

$=\sqrt{72}=6 \sqrt{2}$

Hence, $A B=B C=A C$

Therefore, Points A, B, C make an equilateral triangle.