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

Asked by Sakshi | 1 year ago |  46

##### Solution :-

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, AB2 + BC2

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

= 18 + 18

= 36

= AC2

We know that, AB = BC and AB2 + BC2 = AC2

So, Δ ABC is an isosceles-right angled triangle

Hence Proved.

Answered by Aaryan | 1 year ago

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