Question

The points $(2a,4a),(2a,6a)$ and $(2a+\sqrt{3}a,5a)$ are the vertices of

• A. an isosceles triangle
• B. an equilateral triangle
• C. a right-angled isosceles triangle
• D. a scalene triangle

Answer 1

The correct option is B an equilateral triangle

Let $A\equiv (2a,4a),B\equiv (2a,6a)$ and $C\equiv (2a+\sqrt{3}a,5a)$
Now, $AB=\sqrt{(2a-2a{)}^2+(6a-4a{)}^2}=\sqrt{4a^2}=2|a|$
$BC=\sqrt{(2a+\sqrt{3}a-2a{)}^2+(5a-6a{)}^2}=\sqrt{4a^2}=2|a|$
and $AC=\sqrt{(2a+\sqrt{3}a-2a{)}^2+(5a-4a{)}^2}=\sqrt{4a^2}=2|a|$
Since, $AB=BC=CA$, hence points $A,B$ and $C$ are the vertices of an equilateral triangle.