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 an isosceles triangle Isosceles right-angled triangle is a triangle whose two sides are equal
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
Clearly,
AB = BC
Thus, Δ ABC is an isosceles triangle
Hence Proved
