Question

the coordinates of the third vertex of an equilateral triangle whose two vertices are at (3, 4) and (-2, 3) are

• A. (1, 1) or (1, -1)
• B. $(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2})$ or $(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2})$
• C. (0, 1)
• D. (1, 10)

Answer 1

Answer: (B)

$(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2})$ or $(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2})$

Let the vertices of the equilateral triangle be $A,B$ and $C$

Let $A=(3,4)$,$B=(-2,3)$ and $C=(x,y)$

Since $ABC$ is an equilateral triangle,$AC=BC=AB$

By distance formula,

$AC=\sqrt{{(x-3)}^2+{(y-4)}^2}=\sqrt{x^2-6x+y^2-8y+25}$

$BC=\sqrt{{(x+2)}^2+{(y-3)}^2}=\sqrt{x^2+4x+y^2-6y+13}$

$AB=\sqrt{{(-2-3)}^2+{(3-4)}^2}=\sqrt{26}$

Now ,$AC=BC$

$\Rightarrow \sqrt{x^2-6x+y^2-8y+25}=\sqrt{x^2+4x+y^2-6y+13}$

Squaring both sides, we get

$\Rightarrow x^2-6x+y^2-8y+25=x^2+4x+y^2-6y+13$

$\Rightarrow -6x-8y+25-4x+6y-13=0$

$\Rightarrow -10x-2y+12=0$

$\Rightarrow 5x+y=6$

$\Rightarrow y=6-5x$

Also,$AC=AB$

$\Rightarrow \sqrt{x^2-6x+y^2-8y+25}=\sqrt{26}$

$\Rightarrow x^2-6x+y^2-8y+25=26$

$\Rightarrow x^2+y^2-6x-8y=1$

Substituting $y=6-5x$ in the above equation, we get

$\Rightarrow x^2+{(6-5x)}^2-6x-8(6-5x)=1$

$\Rightarrow x^2+36+25x^2-60x-6x-48+40x=1$

$\Rightarrow 26x^2-26x-13=0$

$\Rightarrow 2x^2-2x-1=0$

Solving the above quadratic equation, we get

$x=\frac{2\pm \sqrt{4-4\times 2\times -1}}{4}=\frac{2\pm 2\sqrt{3}}{4}=\frac{1\pm \sqrt{3}}{2}$

$y=6-5x=6-5\left(\frac{1\pm \sqrt{3}}{2}\right)=\frac{12-5\pm 5\sqrt{3}}{2}=\frac{7\pm \sqrt{3}}{2}$

Hence, the coordinates of the third vertex is $(\frac{1+\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2})$ or $(\frac{1-\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2})$