Question

prove that points (2a,4a), (2a,6a), and (2a +√3a,5a) are the vertices of an equilateral triangle of side 2a

Answer 1

hope it will help you

Answer 2

Solution:

Let A(2a,4a),B(2a,6a) and C(2a+√3a,5a) are veriticies of

a triangle ABC.

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Distance formula:

The distance between the

points (x1,y1) and (x2,y2) is

√(x2-x1)²+(y2-y1)²

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i )A(2a,4a), B(2a,6a)

Distance AB = √(2a-2a)²+(6a-4a)²

= √(2a)²

= 2a ----(1)

ii) B(2a,6a) and C(2a+√3a,5a)

Distance BC

= √(2a+√3a-2a)²+(5a-6a)²

= √(√3a)²+(-a)²

= √3a²+a²

= √4a²

= √(2a)²

= 2a ---(2)

iii) C(2a+√3a,5a) and A(2a,4a)

Distance CA

= √[2a-(2a-√3a)]²+(4a-5a)²

= √(2a-2a+√3a)²+(-a)²

= √(√3a)²+a²

=√3a²+a²

= √4a²

= √(2a)²

= 2a

Therefore,

AB = BC = CA = 2a

∆ABC is a equilateral triangle.

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