Question

Find the third vertex of an equilateral triangle whose other two vertices are (1,1) and (-1,-1) respectively is 3

• A. $\left(\sqrt{3},-\sqrt{3}\right)$
• B. $\left(-\sqrt{3},\sqrt{3}\right)$
• C. $\left(-\sqrt{3},-\sqrt{3}\right)$
• D. Both 1 and 2

Answer 1

The correct option is D Both 1 and 2

Given $△ABC$ is an equilateral triangle

Let the third coordinates be $(a,b)$

Now, $AB=AC$

$\therefore \sqrt{(a+1{)}^2+(b+1{)}^2}=\sqrt{(a-1{)}^2+(b-1{)}^2}$

$\Rightarrow (a^2+1{)}^2+(b+1{)}^2=(a-1{)}^2+(b-1{)}^2$

$\Rightarrow a^2+2a+1+b^2+1+2b=a^2+1-2a+b^2+1-2b$

$\Rightarrow 2a+2b=-2a-2b$

$\Rightarrow 4a+4b=0$

$\Rightarrow a=-b$

Now $B{C}^2=B=A{B}^2=(a+1{)}^2+(b+1{)}^2$

$8=a^2+b^2+2a+2b+2$

Putting $a=-b$

we egt $b=\pm \sqrt{3}$

$\therefore A=(\sqrt{3},-\sqrt{3})$ or $(-\sqrt{3},\sqrt{3}$