Question

an equilateral triangle has each side equal to a . If the co-ordinates of the vertices are ( x1 , y1 ),( x2 , y2 )and (x3 , y3) show that : | x1 y1 1 x2 y2 1 x3 y3 1 |^2

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{ABC}\mathrm{be}\mathrm{an}\mathrm{equilateral}∆\mathrm{having}\mathrm{its}\mathrm{vertices}\mathrm{at}\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right);\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right);\mathrm{C}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right) \\ \mathrm{Let}∆=\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right| \\ \Rightarrow ∆={\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)-{\mathrm{y}}_1\left({\mathrm{x}}_2-{\mathrm{x}}_3\right)+1\left({\mathrm{x}}_2{\mathrm{y}}_3-{\mathrm{x}}_3{\mathrm{y}}_2\right) \\ \Rightarrow ∆={\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)-{\mathrm{x}}_2{\mathrm{y}}_1+{\mathrm{x}}_2{\mathrm{y}}_3+{\mathrm{x}}_3{\mathrm{y}}_1-{\mathrm{x}}_3{\mathrm{y}}_2 \\ \Rightarrow ∆={\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right) \\ \Rightarrow ∆=2\times \frac{1}{2}\left[{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right] \\ \Rightarrow ∆=2\times \mathrm{area}\mathrm{of}∆\mathrm{ABC} \\ \Rightarrow ∆=2\times \frac{\sqrt{3}}{4}{\mathrm{a}}^2 \\ \Rightarrow ∆=\frac{\sqrt{3}}{2}{\mathrm{a}}^2 \end{gathered}$$