# Show that the points P

Question:

Show that the points P(1, 3, 4), Q(-1, 6, 10), R(-7, 4, 7) and S(-5, 1, 1) are the vertices of a rhombus.

Solution:

To prove: Points P, Q, R, S forms rhombus.

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,

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

$\left(x_{2}, y_{2}, z_{2}\right)=(-1,6,10)$

$\left(x_{3}, y_{3}, z_{3}\right)=(-7,4,7)$

$\left(x_{4}, y_{4}, z_{4}\right)=(-5,1,1)$

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

$=\sqrt{(-1-1)^{2}+(6-3)^{2}+(10-4)^{2}}$

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

$=\sqrt{4+9+36}$

$=\sqrt{49}$

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

$=\sqrt{(-7+1)^{2}+(4-6)^{2}+(7-10)^{2}}$

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

$=\sqrt{36+34+9}$

$=\sqrt{49}$

Length RS $=\sqrt{\left(\mathrm{x}_{4}-\mathrm{x}_{3}\right)^{2}+\left(\mathrm{y}_{4}-\mathrm{y}_{3}\right)^{2}+\left(\mathrm{z}_{4}-\mathrm{z}_{3}\right)^{2}}$

$=\sqrt{(-5+7)^{2}+(1-4)^{2}+(1-7)^{2}}$

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

$=\sqrt{4+9+36}$

$=\sqrt{49}$

Length $P S=\sqrt{\left(x_{4}-x_{1}\right)^{2}+\left(y_{4}-y_{1}\right)^{2}+\left(z_{4}-z_{1}\right)^{2}}$

$=\sqrt{(-5-1)^{2}+(1-3)^{2}+(1-4)^{2}}$

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

$=\sqrt{36+4+9}$

$=\sqrt{49}$

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

$=\sqrt{(-7-1)^{2}+(4-3)^{2}+(7-4)^{2}}$

$=\sqrt{(-8)^{2}+(1)^{2}+(3)^{2}}$

$=\sqrt{64+1+9}$

$=\sqrt{74}$

Length $Q S=\sqrt{\left(x_{4}-x_{2}\right)^{2}+\left(y_{4}-y_{2}\right)^{2}+\left(z_{4}-z_{2}\right)^{2}}$

$=\sqrt{(-5+1)^{2}+(1-6)^{2}+(1-10)^{2}}$

$=\sqrt{(-4)^{2}+(-5)^{2}+(-9)^{2}}$

$=\sqrt{16+25+81}$

$=\sqrt{122}$

Here, $P Q=R S=Q R=P S$.

Also the diagonals $P R \neq Q S$.

Hence, the polygon is a rhombus as all sides are equal and diagonals are not equal.