The points (a,a),(–a,−a) and (–√3a,√3a) are t

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Question

The points

(a, a), (- a, - a)

and

(- \sqrt{3} a, \sqrt{3} a)

are the vertices of a/an _____ triangle.

Options:

A . scalene

B . right angled

C . isosceles right angled

D . equilateral

Answer: Option D
:
D
Let's name the points as

A (a, a), B (- a, - a)

and

C (- \sqrt{3} a, \sqrt{3} a)

.
By distance formula, distance between points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

B C = \sqrt{(- \sqrt{3} a + a)^{2} + (\sqrt{3} a + a)^{2}} = \sqrt{a^{2} + (\sqrt{3} a)^{2} - 2 \times a \times \sqrt{3} a + \sqrt{3} a)^{2} + a^{2} + 2 \times a \times \sqrt{3} a} B C = \sqrt{8 a^{2}} = 2 \sqrt{2} a units Similarly A B = \sqrt{(a + a)^{2} + (a + a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units A C = \sqrt{(- \sqrt{3} a - a)^{2} + (\sqrt{3} a - a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units

Weget, A B = B C = A C = 2 \sqrt{2} a units ∴ △ ABC is an equilateral triangle .

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