Question

An equilateral triangle has two vertices at the points (3, 4) and (-2, 3), find the coordinates of the third vertex.

Answer 1

Sol:
Two vertices of an equilateral triangle are (3, 4) and (-2, 3)
Let the third vertex of the triangle be (x, y)
Distance between (3, 4) and (-2, 3)
=$\sqrt{(-2-3{)}^2+(3-4{)}^2}$
=$\sqrt{(-5{)}^2+(-1{)}^2}$
⇒ $\sqrt{26}$

Distance between (3, 4) and (x, y)
= $\sqrt{\left[\right(x-3{)}^2+(y-4{)}^2]}=\sqrt{x^2-6x+9+y^2-8y+16}=\sqrt{x^2-6x+y^2-8y+25}$ ------------------------ (1)

Distance between (-2, 3) and (x, y)
= $\sqrt{(x+2{)}^2+(y-3{)}^2}=\sqrt{(x+2{)}^2+(y-3{)}^2}=\sqrt{x^2+4x+4+y^2-6y+9}=\sqrt{x^2+4x+y^2-6y+13}$-------------------- (2)

Equating the distances we get,
$x^2-6x+y^2-8y+25=x^2+4x+y^2-6y+13$
10x + 2y - 12 = 0
5x + y - 6 = 0
y = (6 - 5x)

Substituting the value of y in equation (1) and equating it to $\sqrt{26}$ snd squaring on both sides we get

$x^2-6x+y^2-8y+25=26\Rightarrow x^2-6x+(6-5x{)}^2-8(6-5x)+25=26\Rightarrow x^2-6x+36+25x^2-60x-48+40x+25=26\Rightarrow 26x^2-26x-13=0\Rightarrow 2x^2-2x-1=0$

Solving the quadratic equation using the quadratic formula, $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

x = $\frac{2\pm \sqrt{4+8}}{4}$
x = $\frac{2\pm \sqrt{12}}{4}$
x = $\frac{2\pm 2\sqrt{3}}{4}$
x = $\frac{1\pm \sqrt{3}}{2}$

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

Hence, the coordinates of the third vertex of the equilateral triangle are $\frac{[1\pm \sqrt{3}]}{2},\frac{7\pm 5\sqrt{3}}{2}$.