Question

The triangle formed by the points $A(2a,4a),B(2a,6a)$and $C(2a+\sqrt{3}a,5a)$ is a/an ____.

• A. right angled triangle
• B. isosceles triangle
• C. equilateral triangle
• D. None of these

Answer 1

The correct option is C equilateral triangle

The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{{(2a-2a)}^2+{(6a-4a)}^2}=\sqrt{0+4a^2}=2a$
$BC=\sqrt{{(2a+\sqrt{3}a-2a)}^2+{(5a-6a)}^2}=\sqrt{3a^2+a^2}=2a$
And $AC=\sqrt{{(2a+\sqrt{3}a-2a)}^2+{(5a-4a)}^2}=\sqrt{3a^2+a^2}=2a$
Thus, $AB=BC=CA$

Hence $ABC$ is an equilateral triangle.