Question

Prove that the coordinate of the third vertex of an equilateral triangle whose two vertices are (3,4) and (-2,3) are{(1+√3)/2 , (7-5√3)/2} or {(1-√3)/2 ,(7+5√3)/2}

Answer 1

$$\begin{gathered} \mathrm{Let}ABC\mathrm{be}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}. \\ \therefore AB=BC=CA \\ \Rightarrow A{B}^2=B{C}^2=C{A}^2 \\ \mathrm{Let}A\equiv \left(3,4\right) \\ B\equiv \left(-2,3\right) \\ \mathrm{And}C\equiv \left(x,y\right) \\ AB=\sqrt{{\left(3+2\right)}^2+{\left(4-3\right)}^2} \\ =\sqrt{5^2+1^2}=\sqrt{26} \\ \mathrm{Now}, \\ A{C}^2=B{C}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-4\right)}^2={\left(x+2\right)}^2+{\left(y-3\right)}^2 \\ \Rightarrow x^2-6x+9+y^2-8y+16=x^2+4x+4+y^2-6y+9 \\ \Rightarrow -6x+9-8y+16=4x+4-6y+9 \\ \Rightarrow -10x-2y+12=0 \\ \Rightarrow 5x+y-6=0 \\ \Rightarrow 5x+y-6=0 \\ \Rightarrow y=6-5x.....\left(1\right) \\ \mathrm{Now}, \\ A{C}^2=A{B}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-4\right)}^2=26 \\ \Rightarrow x^2-6x+9+y^2-8y+16=26 \\ \Rightarrow x^2-6x+y^2-8y=1 \\ \Rightarrow x^2-6x+{\left(6-5x\right)}^2-8\left(6-5x\right)=1\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow x^2-6x+36-60x+25x^2-48+40x=1 \\ \Rightarrow 26x^2-26x-13=0 \\ \Rightarrow 2x^2-2x-1=0 \\ \Rightarrow x=\frac{--2\pm \sqrt{{\left(-2\right)}^2-4\times \left(2\right)\times -1}}{2\times 2} \\ =\frac{2\pm \sqrt{4+8}}{4} \\ =\frac{2\pm \sqrt{12}}{4} \\ =\frac{2\pm 2\sqrt{3}}{4} \\ =\frac{1\pm \sqrt{3}}{2} \\ \mathrm{For}x=\frac{1+\sqrt{3}}{2} \\ y=6-5\times \frac{1+\sqrt{3}}{2}=\frac{12-5-5\sqrt{3}}{2}=\frac{7-5\sqrt{3}}{2} \\ \mathrm{And} \\ \mathrm{For}x=\frac{1-\sqrt{3}}{2} \\ y=6-5\times \frac{1-\sqrt{3}}{2}=\frac{12-5+5\sqrt{3}}{2}=\frac{7+5\sqrt{3}}{2} \\ \therefore C\equiv \left(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2}\right)or\left(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2}\right) \end{gathered}$$