Ortho centre of a triangle ABC is H(2,3) and altitude AH meets the circum circle of triangle ABC at D(4, 7) then

H is ortho centre of $\Delta ABC$

            $BC\perp HD$

            Slope of HD=2

            $\perp lr$ line slope $=-\frac{1}{2}$

            Mid point of HD=(3, 5)

            Equation of BC is

            $y-5=-\frac{1}{2}(x-3)$

            $x+2y=13$

            $HD=SD=2\sqrt{5}=R$

            Of ABC is an equilateral triangle

            $r=\frac{R}{2}=\sqrt{5}$

            $A=2H-D=(0,-1)$

            Area of equilateral triangle $ABC=\frac{\sqrt{3}}{4}{\left(side\right)}^2$

            $\because highth=3\sqrt{5}=\frac{\sqrt{3}}{4}\times \frac{4}{3}\times {\left(3\sqrt{5}\right)}^2$

            Side $=\frac{2}{\sqrt{3}}h=\frac{1}{\sqrt{3}}\times 9\times 5$

            $\therefore Area=15\sqrt{3}=\frac{3\times \sqrt{3}\times \sqrt{3}5}{\sqrt{3}}$

                                                $=15\sqrt{3}$