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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Solution

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

AB = $\sqrt{(0 + 1)^{2} + (7 - 6)^{2} + (10 - 6)^{2}}$

$= \sqrt{(1)^{2} + (1)^{2} + (4)^{2}}$

$= \sqrt{18}$ $= 3 \sqrt{2}$

$A C = \sqrt{(0 + 4)^{2} + (7 - 9)^{2} + (10 - 6)^{2}}$

$= \sqrt{(4)^{2} + (- 2)^{2} + (4)^{2}}$

$= \sqrt{36}$ $(A B)^{2} + (B C)^{2} = A C^{2}$

$\Rightarrow ∠ A B C = 90^{\circ}$ and AB = BC

Hence, (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an

isosceles right - angled triangle.

Suggest Corrections

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If the points $(0, 7, 10), (- 1, 6, 6) a n d (- 4, 9, 6)$ are the vertices of a triangle, then the triangle is _____

Verify the following:

(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle.

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

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