Given:

The points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6)

Isosceles right-angled triangle is a triangle whose two sides are equal and also satisfies Pythagoras Theorem.

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

Since, AB = BC

So, AB^{2} + BC^{2}

= \( (3\sqrt{2})^2 + (3\sqrt{2})^2\)

= 18 + 18

= 36

= AC^{2}

We know that, AB = BC and AB^{2} + BC^{2} = AC^{2}

So, Δ ABC is an isosceles-right angled triangle

Hence Proved.

Answered by Aaryan | 1 year agoA(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

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