Question

verify that (0,7,10),(-1,6,6)and(-4,9,6) are the vertices of a right angled triangle.​

Answer 1

Given :

The vertices of a right angled triangle are,

•A=(0,7,10)

•B=(-1,6,6)

•C=(-4,9,6)

To find :

•To verify that the given points are the vertices of a right angled triangle =?

Formula :

•Distance Formula :

$\pink\bigstar\boxed{\bf{\sqrt{(y_{2}-y_{1} )^{2}+(x_{2}-x_{1})^{2}+(z_{2}-z_{1})^{2}}}}$

Solution :

Let (0,7,10) ,(-1,6,6),(-4,9,6) be denoted by A, B, and C respectively.

$\\ \implies\displaystyle\sf{AB=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2} +(z_{2}-z_{1})^{2}}}$

$\\ \implies\displaystyle\sf{AB=\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2} }}$

$\\ \implies\displaystyle\sf{AB=\sqrt{(-1)^{2}+(-1)^{2}+(-4)^{2}}}$

$\\ \implies\displaystyle\sf{AB=\sqrt{1+1+16}}$

$\\ \implies\displaystyle\sf{AB=\sqrt{18}}$

$\\ \implies\boxed{\displaystyle\sf{AB=3\sqrt{2} }}$

$\\ \implies\displaystyle\sf{BC=\sqrt{(z_{2}-z_{1})^{2}+(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}}$

$\\ \implies\displaystyle\sf{BC=\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2} }}$

$\\ \implies\displaystyle\sf{BC=\sqrt{(-3)^{2}+(3)^{2} }}$

$\\ \implies\displaystyle\sf{BC=\sqrt{9+9}}$

$\\ \implies\boxed{\displaystyle\sf{BC=3 \sqrt{2} }}$

$\\ \implies\displaystyle\sf{CA=\sqrt{(z_{2}-z_{1})^{2}+(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2} }}$

$\\ \implies\displaystyle\sf{CA=\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}}$

$\\ \implies\displaystyle\sf{CA=\sqrt{(4)^{2}+(-2)^{2}+(4)^{2} }}$

$\\ \implies\displaystyle\sf{CA=\sqrt{16+4+16}}$

$\\ \implies\displaystyle\sf{CA=\sqrt{36}}$

$\\ \implies\boxed{\displaystyle\sf{CA=6}}$

Now,

$\\ \implies\displaystyle\sf{AB^{2}+BC^{2}=(3\sqrt 2)^{2}+(3 \sqrt 2)^{2} }$

$\\ \implies\displaystyle\sf{18+18=36=AC^{2} }$

Therefore, by Pythagoras theorem ,ABC is a right angled triangle .

Hence ,the given points are the vertices of a right angled triangle .