• # question_answer If $A({{x}_{1}},\,\,{{y}_{1}}),$ $B({{x}_{2}},\,\,{{y}_{2}})$ and $C({{x}_{3}},\,\,{{y}_{3}})$ are vertices of an equilateral triangle whose each side is equal to a, then prove that

Given, $A\,({{x}_{1}},\,\,{{y}_{1}}),$ $B\,({{x}_{2}},\,\,{{y}_{2}})$ and $C\,({{x}_{3}},\,\,{{y}_{3}})$are vertices of an equilateral triangle and each side of equilateral triangle = a Now, LHS = By taking common 2 from ${{C}_{3}}$ in both the determinant, we get $=4\times (2\times Area\,\,of\,\,equilateral\,\,\Delta ABC)$ $\times \,(2\times \text{Area}\,\,\text{of}\,\,\text{equilateral}\,\,\Delta ABC)$ [$\because$area of triangle whose vertices are $({{x}_{1}},\,\,{{y}_{1}}),\,\,({{x}_{2}},\,\,{{y}_{2}})$and $=4\times 4\times {{[Area\,\,of\,\,equilateral\,\,\Delta \,ABC]}^{2}}$ $=16\times {{\left[ \frac{\sqrt{3}}{4}\times {{a}^{2}} \right]}^{2}}$ $=16\times \frac{3}{16}\times {{a}^{4}}$ [$\because$area of equilateral triangle$=\frac{\sqrt{3}}{4}{{(side)}^{2}}$] $=3{{a}^{4}}=RHS$             Hence proved.