The triangle formed by the points (0,7,10), (- 1,6,6), (- 4,9,6) is

A
equilateral
B
isosceles
C
right angled
D
right angled isosceles

Solution:

Let A(0,7,10), B(-1,6,6) and C(-4 ,9,6)

Then,

$\begin{matrix} AB = \sqrt{(- 1 - 0)^{2} + (6 - 7)^{2} + (6 - 10)^{2}} \\ = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2} \\ BC = \sqrt{(- 4 + 1)^{2} + (9 - 6)^{2} + (6 - 6)^{2}} \\ = \sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2} \\ AC = \sqrt{(- 4 - 0)^{2} + (9 - 7)^{2} + (6 - 10)^{2}} \\ = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \\ Clearly, A C^{2} = {AB}^{2} + {BC}^{2} \end{matrix}$

Hence, triangle is right angled. Also, AB = BC

Triangle is right angled isosceles.

Solution:

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