Q201 / 270 · NDA 2016 (Phase 2) · Mathematics

An equilateral triangle has one vertex at $(0,0)$ and another at $(3, \sqrt{3})$. What are the coordinates of the third vertex?

  • A
    $(0,2 \sqrt{3})$ only
  • B
    $(3,-\sqrt{3})$ only
  • Correct answer
    C
    $(0,2 \sqrt{3})$ or $(3,-\sqrt{3})$
  • D
    Neither $(0,2 \sqrt{3})$ nor $(3,-\sqrt{3})$

Solution

Let $A B C$ is equilateral triangle with $A(0,0)$ and $B(3, \sqrt{3})$ and $C$ to be known. $\therefore A B=\sqrt{(3-0)^2+(\sqrt{3}-0)^2}=\sqrt{9+3}=\sqrt{12}$ Take option (a) i.e. $C(0,2 \sqrt{3})$ $\begin{aligned} & C A=\sqrt{0^2+(2 \sqrt{3})^2}=\sqrt{12} \\ & C B=\sqrt{(3)^2+(\sqrt{3})^2}=\sqrt{12} \end{aligned}$ Take option (b) i.e. $C(3,-\sqrt{3})$ $\begin{aligned} & C A=\sqrt{3^2+(\sqrt{3})^2}=\sqrt{12} \\ & C B=\sqrt{(0)^2+(2 \sqrt{3})^2}=\sqrt{12} \end{aligned}$ $\therefore$ Both option (1) and (2) are correct.

Hence, the correct answer is option: 'C'.

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