Question

(2a, 4a), (2a, 6a) and (2a +
$\sqrt{3a}$
, 5a) are vertices
of a/an
(1) Isosceles triangle
(2) Scalene triangle
(3) Equilateral triangle
(4) Isosceles right triangle​

Answer 1

Distance Formula

→ √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(2a - 2a)² + (6a - 4a)²] = 2a

BC = √[(2a - 2a)² + (6a - √3a)²] = 6a - √3a

CA = √[(2a - 2a)² + (4a - √3a)²] = 4a - √3a

Since AB, BC and CA both have unequal length.

Hence this triangle is Scalene Triangle

Answer 2

Answer:-

Refer the attachment