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

Equilateral

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Isosceles

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Right angled

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Right angled Isosceles

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Solution

The correct option is D
Right angled Isosceles

Let A be (0, 7, 10), B (–1, 6, 6) and C (– 4, 9, 6)
Then, AB = $\sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$
BC = $\sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2}$
AC = $\sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Clearly, $A C^{2} = A B^{2} + B C^{2}$
Hence $△$ is right angled. Also, since AB = BC
$∴ △$ is right angled isosceles.

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Similar questions

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