The points a -3- a -2 a -3- 2 a - a -3- 3 a are the vertices ofA- An equilateral triangleB- An isosceles triangleC- A right angled triangleD- None of these

The correct option is B An isosceles triangle Let A( a/√(3), a ), B( 2a/√(3), 2a ) and C( a/√(3), 3a ) Then AB^2 = (a/√(3) 2a/√(3))^2 + (a 2a)^2 = a^2/ ...

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Byju's Answer

Standard XI

Mathematics

Distance formula using Complex Numbers

The points a ...

Question

The points $(\frac{a}{\sqrt{3}}, a), (\frac{2 a}{\sqrt{3}}, 2 a), (\frac{a}{\sqrt{3}}, 3 a)$ are the vertices of

An equilateral triangle

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An isosceles triangle

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A right angled triangle

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None of these

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Solution

The correct option is B
An isosceles triangle

Let A $(\frac{a}{\sqrt{3}}, a), B (\frac{2 a}{\sqrt{3}}, 2 a) a n d C (\frac{a}{\sqrt{3}}, 3 a)$

Then $A B^{2}$ = ${(\frac{a}{\sqrt{3}} - \frac{2 a}{\sqrt{3}})}^{2}$ + $(a - 2 a)^{2}$ = $\frac{a^{2}}{3}$ + $a^{2}$ = $\frac{4 a^{2}}{3}$

Similarly $B C^{2}$ = $\frac{4 a^{2}}{3}$ and $A C^{2}$ = $4 a^{2}$

Hence it is an isosceles triangle.

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The points (a√3,a),(2a√3,2a),(a√3,3a) are the vertices of

The correct option is B An isosceles triangle Let A(a√3,a),B(2a√3,2a)and C(a√3,3a) Then AB2 = (a√3−2a√3)2 + (a−2a)2 = a23 + a2 = 4a23 Similarly BC2 = 4a23 and AC2 = 4a2 Hence it is an isosceles triangle.

The points $(\frac{a}{\sqrt{3}}, a), (\frac{2 a}{\sqrt{3}}, 2 a), (\frac{a}{\sqrt{3}}, 3 a)$ are the vertices of