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/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

An isosceles triangle

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A right angled triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

A (4, 4), B (3, 5)

and

C (- 1, - 1)

are the vertices of

The following points A(2a, 4a), B(2a, 6a) and C $(2 a + \sqrt{3} a, 5 a)$ , (a > 0) are the vertices of

Q. The points

(2 a, 4 a), (2 a, 6 a)

and

(2 a + \sqrt{3} a, 5 a)

are the vertices of

The points (0, 0), (a, 0) and $(\frac{a}{2}, \frac{a \sqrt{3}}{2})$ are vertices of

The points A (1, 1), B (-1, 1) and the origin form the vertices of

Join BYJU'S Learning Program

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