Show that the points A(0, 7, 10), B(-1, 6, 6) and C(-4, 9,6) form an isosceles right-angled triangle.

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Solution

We have

$A B = \sqrt{(- 1 - 0)^{2} + (6 - 7)^{2} + (6 - 10)^{2}} = \sqrt{18} = 3 \sqrt{2}$

$B C = \sqrt{(- 4 + 1)^{2} + (9 - 6)^{2} + (6 - 6)^{2}} = \sqrt{18} = 3 \sqrt{2}$

and $A C = \sqrt{(- 4 - 0)^{2} + (9 - 7)^{2} + (6 - 7)^{2}} = \sqrt{36} = 6$ .

Clearly, AB=BC and $A B^{2} + B C^{2} = (18 + 18) = 36 = A C^{2}$

Hence, triangle ABC is an isosceles right-angled triangle.

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

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The triangle formed by the points (0, 7, 10), (–1, 6, 6),(– 4, 9, 6) is [RPET 2001]

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