Question

Verify the following: 

(0, 7, 10), (-1, 6, 6) and (2, -3, 4) are vertices of a right-angled triangle

Answer 1

Given: Points are A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) 

To prove: the triangle formed by given points is a right-angled triangle Right-angled triangle satisfies Pythagoras Theorem 

Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by,

  \(\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}\)

Therefore, 

Distance between A(0, 7, 10) and B(-1, 6, 6) is AB,

= \(\sqrt{(0-(-1))^2+(7-6)^2+(10-6)^2}\) 

= \(\sqrt{(1)^2+1^2+(-4)^2}\) 

= \(\sqrt{1+1+16}\) 

= \(\sqrt{18}\) 

= \(3\sqrt{2}\) 

Distance between B(-1, 6, 6) and C(-4, 9, -6) is BC,

= \(\sqrt{(-1-(-4))^2+(6-9)^2+(6-6)^2}\) 

= \(\sqrt{3^2+(-3)^2+0^2}\) 

= \(\sqrt{9+9}\) 

= \(\sqrt{18}\) 

= \(3\sqrt{2}\) 

Distance between A(0, 7,10) and C(-4, 9, 6) is AC,

= \(\sqrt{(0-(-4))^2+(7-9)^2+(10-6)^2}\) 

= \(\sqrt{4^2+(-2)^2+4^2}\) 

= \(\sqrt{16+4+16}\) 

= \(\sqrt{36}\) 

= 6

AB² + BC²

= \((3\sqrt{2})^2 +(3\sqrt{2})^2 \) 

= 18+18

= 36

=AC²

As, AB² + BC² = AC² 

Thus, Δ ABC is a right angled triangle 

Hence Proved