Question

$ABC$ is an equilateral triangle. If the coordinates of two of its vertices are ($1,3)$ and $(-2,7)$ the coordinates of the third vertex can be

• A. $(-\frac{1}{2}+2\sqrt{3},5+\frac{3\sqrt{3}}{2})$
• B. $(-\frac{1}{2}+2\sqrt{3},5-\frac{3\sqrt{3}}{2})$
• C. $(\frac{1}{2}-2\sqrt{3},-5+\frac{3\sqrt{3}}{2})$
• D. $(\frac{1}{2}+2\sqrt{3},-5-\frac{3\sqrt{3}}{2})$

Answer 1

Answer: (A), (B)

The correct options are
A $(-\frac{1}{2}+2\sqrt{3},5+\frac{3\sqrt{3}}{2})$
B $(-\frac{1}{2}+2\sqrt{3},5-\frac{3\sqrt{3}}{2})$

Let $A\equiv (1,3),B\equiv (-2,7)$ and $C\equiv (x,y)$
As $△ABC$ is equilateral then $AB=AC=BC$
$AB=AC\Rightarrow {\left(AB\right)}^2={\left(AC\right)}^2\Rightarrow {(-2-1)}^2+{(7-3)}^2={(x-1)}^2+{(y-3)}^2$
$\Rightarrow x^2+y^2-2x-6y-15=0$ ...(1)
And
$AB=BC\Rightarrow {\left(AB\right)}^2={\left(BC\right)}^2\Rightarrow {(-2-1)}^2+{(7-3)}^2={(x+2)}^2+{(y-7)}^2$
$\Rightarrow x^2+y^2+4x-14y+29=0$ ...(2)
Solving (1) and (2)
$(x,y)=(-\frac{1}{2}+2\sqrt{3},5\pm \frac{3\sqrt{3}}{2})$