Question

Verify the following: 

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

Answer 1

Let A(0,7,10) , B( \[-\]1,6,6) and C( \[-\]4,9,6) be the vertices of \[\bigtriangleup ABC\]Then ,

AB = \[\sqrt{\left( - 1 - 0 \right)^2 + \left( 6 - 7 \right)^2 + \left( 6 - 10 \right)^2}\]

\[= \sqrt{\left( - 1 \right)^2 + \left( - 1 \right)^2 + \left( - 4 \right)^2}\]
\[ = \sqrt{1 + 1 + 16}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

BC = \[\sqrt{\left( - 4 + 1 \right)^2 + \left( 9 - 6 \right)^2 + \left( 6 - 6 \right)^2}\]

\[= \sqrt{\left( - 3 \right)^2 + \left( 3 \right)^2 + 0}\]
\[ = \sqrt{9 + 9}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

AC = \[\sqrt{\left( - 4 - 0 \right)^2 + \left( 9 - 7 \right)^2 + \left( 6 - 10 \right)^2}\]

\[= \sqrt{\left( - 4 \right)^2 + \left( 2 \right)^2 + \left( - 4 \right)^2}\]
\[ = \sqrt{16 + 4 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]

\[A C^2\]\[= A B^2 + B C^2\]

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