Question

The points $A(2a,4a)$, $B(2a,6a)$ and $C(2a+\sqrt{3}a,5a)$ (when $a>0$) are vertices of :

• A. an obtues angled triangle
• B. an equilateral triangle
• C. an isosceles obtuse angled triangle
• D. a right angled triangle

Answer 1

The correct option is B an equilateral triangle

Given points are

$A(x_1,y_1)=(2a,4a)$

$B(x_2,y_2)=(2a,6a)$

and $C(x_3,y_3)=(2a+\sqrt{3}a,5a)$

Distance of

$AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

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

$=\sqrt{{\left(2a\right)}^2}$

$AB=2a$

Distance of

$BC=\sqrt{{(x_2-x_3)}^2+{(y_2-y_3)}^2}$

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

$=\sqrt{{\left(\sqrt{3}a\right)}^2+a^2}$

$BC=2a$

Distance of

$CA=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2}$

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

$=\sqrt{{\left(\sqrt{3}a\right)}^2+a^2}$

$CA=2a$

Then $AB=BC=CA=2a$

Hence, it is an equilateral triangle.