Question

If $(2,4),(2,6)$ are two vertices of an equilateral triangle then the third vertex is

• A. $(2+\sqrt{3},5)$
• B. $(\sqrt{3}-2,5)$
• C. $(5,2+\sqrt{3})$
• D. $(5,2-\sqrt{3})$

Answer 1

Answer: (A)

$(2+\sqrt{3},5)$

Let the coordinate of 6 be $(x,y)$ in an equilateral triangle all sides are of equal length.

$\Rightarrow AB=AC=BC$

Here $AB=\sqrt{{(2-2)}^2+{(6-4)}^2}=\sqrt{2^2}=2units$

and $AC=\sqrt{{(2-x)}^2+{(4-y)}^2}=2units........\left(1\right)$

$BC=\sqrt{{(2-x)}^2+{(6-y)}^2}=2units.......\left(2\right)$

also, $AC=BC$

$\Rightarrow \sqrt{{(2-x)}^2+{(4-y)}^2}=\sqrt{{(2-x)}^2+{(6-y)}^2}..........\left(4\right)$

squaring both sides we get,

$\Rightarrow {(2-x)}^2+{(4-y)}^2={(2-x)}^2+{(6-y)}^2$

$\Rightarrow 16+y^2-8y=36+y^2-12y$

$\Rightarrow 4y=20$

$\Rightarrow y=\mathrm{5......}\left(5\right)$

Now, putting $y=5$ in equation $\left(1\right)$ we have,

$\Rightarrow \sqrt{{(2-x)}^2+1^2}=2$

squaring both sides we have,

$\Rightarrow {(2-x)}^2+1=4$

$\Rightarrow {(2-x)}^2=3$

$\Rightarrow 2-x=\pm \sqrt{3}$

$\Rightarrow x=2\pm \sqrt{3}$

$\therefore$ the possible coordinates of $x$ are $(2-\sqrt{3},5)$ and $(2+\sqrt{3},5)$

Hence, the answer is $(2+\sqrt{3},5).$