Question

If $A(a,a),B(-a,-a)$ are two vertices of an equilateral triangle, then its third vertex is

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

Answer 1

Answer: (B)

$(-a\sqrt{3},a\sqrt{3})$

The two vertices of an equilateral triangle is $A\left(a,a\right)$ and $B\left(-a,-a\right)$.

Let the third vertex be $C\left(x,y\right)$.

Since the triangle is an equilateral triangle, then,

$AC=AB=BC$

$\sqrt{{\left(x-a\right)}^2+{\left(y-a\right)}^2}=\sqrt{{\left(-a-a\right)}^2+{\left(-a-a\right)}^2}=\sqrt{{\left(x+a\right)}^2+{\left(y+a\right)}^2}$

$\sqrt{{\left(x-a\right)}^2+{\left(y-a\right)}^2}=\sqrt{{\left(-2a\right)}^2+{\left(-2a\right)}^2}=\sqrt{{\left(x+a\right)}^2+{\left(y+a\right)}^2}$

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

$\sqrt{{\left(x-a\right)}^2+{\left(y-a\right)}^2}=\sqrt{8a^2}=\sqrt{{\left(x+a\right)}^2+{\left(y+a\right)}^2}$

Taking first two equations,

$\sqrt{{\left(x-a\right)}^2+{\left(y-a\right)}^2}=\sqrt{8a^2}$

Squaring both sides,

${\left(x-a\right)}^2+{\left(y-a\right)}^2=8a^2$

$x^2+a^2-2ax+y^2+a^2-2ay=8a^2$

$x^2+y^2-2ax-2ay=6a^2$ (1)

Taking last two equations,

$\sqrt{8a^2}=\sqrt{{\left(x+a\right)}^2+{\left(y+a\right)}^2}$

Squaring both sides,

$8a^2={\left(x+a\right)}^2+{\left(y+a\right)}^2$

$8a^2=x^2+a^2+2ax+y^2+a^2+2ay$

$x^2+y^2+2ax+2ay=6a^2$ (2)

From equation (1) and (2),

$x^2+y^2-2ax-2ay=x^2+y^2+2ax+2ay$

$-4ax=4ay$

$x=-y$

Substituting the value of $x$ in equation (1),

${\left(-y\right)}^2+y^2-2a\left(-y\right)-2ay=6a^2$

$y^2+y^2+2ay-2ay=6a^2$

$2y^2=6a^2$

$y^2=3a^2$

$y=\sqrt{3}a$

Substituting the value of $y$ in $x=-y$, then,

$x=-\sqrt{3}a$

Therefore, the third vertex is $\left(-\sqrt{3}a,\sqrt{3}a\right)$.