Question

The points $A(2a,4a),B(2a,6a)$ and $C(2a+\sqrt{3}a,5a)$, $\left(\text{when}a>o\right)$ vertices of

Answer 1

$\begin{array}{c}A\left(2a,4a\right),B\left(2a,6a\right)andC\left(2a+\sqrt{3}a,5a\right)\\ \Rightarrow Dis\tan ceofAB=\sqrt{{\left(2a-2a\right)}^2+{\left(6a-4a\right)}^2}\\ AB=\sqrt{0+4a^2}=2aunit\\ \Rightarrow BC=\sqrt{{\left(\sqrt{3a}\right)}^2+{\left(-a\right)}^2}\\ \Rightarrow BC=\sqrt{3a^2+a^2}\\ BC=\sqrt{4a^2}=2aunit\\ \Rightarrow AC=\sqrt{\left(\sqrt{3}{a}^2\right)+{\left(a\right)}^2}\\ \Rightarrow AC=\sqrt{3a^2+a^2}\\ \Rightarrow AC=\sqrt{4a^2}=2aunit\\ AB=BC=CA\\ Hence,A,BandCareVerticesofequilateraltriangle\end{array}$