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Verify the following: (0, 7, -10), (1, 6, -6) and (4, 9, -6) are vertices of an isosceles triangle. |
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Answer» 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 |
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