Question

Show that the points (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of an isosceles right-angled triangle. 

Answer 1

Let A(0, 7, 10), B(1, 6, 6), C(\[-\]4, 9, 6) be the coordinates of △ABC
Then, we have:

4, 9, 6) be the coordinates of △ABC
Then, we have:

\[AB = \sqrt{\left( 0 + 1 \right)^2 + \left( 7 - 6 \right)^2 + \left( 10 - 6 \right)^2}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

\[BC = \sqrt{\left( - 1 + 4 \right)^2 + \left( 6 - 9 \right)^2 + \left( 6 - 6 \right)^2}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

\[AC = \sqrt{\left( 0 + 4 \right)^2 + \left( 7 - 9 \right)^2 + \left( 10 - 6 \right)^2}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[\text{ Clearly }, {AB}^2 + {BC}^2 = {AC}^2 \]
\[ \Rightarrow∠ ABC = 90°\] 
\[& AB = BC\] 

Hence △ABC is a right-angled isosceles triangle.