Question

The coordinates of the third vertex of an equilateral triangle whose two vertices are at $(3,4),(-23)$ are ______.

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

Answer 1

The correct option is C $(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2})$ or $(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2})$

Let $A(3,4)$ and $B(-2,3)$ be the given points and $C(x,y)$ be the third vertex of the equilateral triangle $△$ABC.Then,

$AB=BC=CA$

$⟹A{B}^2=B{C}^2=C{A}^2{[3-(-20\left)\right]}^2+(4-3{)}^2={[x-(-2\left)\right]}^2+(y-3{)}^2=(x-3{)}^2+(y-4{)}^2{5}^2+1^2=(x+2{)}^2+(y-3{)}^2=(x-3{)}^2+(y-4{)}^2⟹26=x^2+2x+4+y^2-6y+9=x^2-6x+9+y^2-8y+16⟹4x-6y+13=-6x-8y+25⟹10x+2y=12\therefore 5x+y=6⟹y=6-5x⟶\left(1\right)x^2+y^2+4x-6y+13=26⟹x^2+(-6-5x{)}^2+4x-6(6-5x)-13=0\therefore x^2+36+25x^2-60x+4x-36+30x-13=0\therefore 26x^2-26x-13=0⟹2x^2-2x-1=0\therefore x=\frac{2\pm \sqrt{12}}{4}=\frac{1+\sqrt{3}}{2}$

when $x=\frac{1+\sqrt{3}}{2}y=6-5\left[\frac{1+\sqrt{3}}{2}\right]=\frac{12-5-5\sqrt{3}}{2}=\frac{7-5\sqrt{3}}{2}$

$whenx=\frac{1-\sqrt{3}}{2},wegety=6-5\left(\frac{1-\sqrt{3}}{2}\right)=\frac{7+5\sqrt{3}}{2}$

$\therefore$ The coordinates are

$[\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2}]or[\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2}]$