Question

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

Answer 1

Answer:

Let A (2a, 4a)

     B(2a, 6a)

    C (2a + sqrt (3a), 5a)

Side AC:

$\sqrt{(2a+\sqrt{3a}-2a)^2+(5a-4a)^2 }=\sqrt{3a+a^2}$

Side BC:

$\sqrt{(2a+\sqrt{3a}-2a)^2+(5a-6a)^2 }=\sqrt{3a+a^2}$

Hence AC = BC

It is an isoceles triangle.

Side AB:

$\sqrt{(2a-2a)^2+(6a-4a)^2 }=\sqrt{2a^2}=2a$

It proves that the triangle is NOT an equilateral triangle.