The line $2x+3y=12$ meets the$x-$axis at $A$ and $y-$axis at $B$. The line through $(5,5)$ perpendicular to $AB$ meets the $x-$axis and the line $AB$ at $C$ and $E$ respectively. If $O$ is the origin, find the area of figure $OCEB$.

Vertices of a triangle are $A\left(x_1,x_1\tan {\alpha }_1\right),B\left(x_2,x_2\tan {\alpha }_2\right)$ and $C\left(x_3,x_3\tan {\alpha }_3\right).$ If the circumcentre coincides with the origin and the orthocentre $H$ is $\left(\overline{x},\overline{y}\right),$ then prove that $\overline{y}\left(\cos {\alpha }_1+\cos {\alpha }_2+\cos {\alpha }_3\right)=\overline{x}\left(\sin {\alpha }_1+\sin {\alpha }_2+\sin {\alpha }_3\right)$ where $x_1{\mathrm{sec\alpha }}_1,x_2{\mathrm{sec\alpha }}_2,x_3{\mathrm{sec\alpha }}_3$ have same sign

The circumcentre of a triangle with vertices $A\left(a\mathit{,}a\tan \alpha \right)\mathit{,}B\left(b\mathit{,}b\tan \beta \right)$ and $C\left(c,c\tan \gamma \right)$ lies at the origin where $\alpha +\beta +\gamma =\pi ,$ show that its orthocentre lies on the line

$4\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}\right)x-4y\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}=y$.

If $p,q,r$ be the lengths of perpendiculars from the vertices $A,B,C$ respectively of the triangle $ABC$ on any straight line, prove that
$a^2\left(p-q\right)\left(p-r\right)+b^2\left(q-r\right)\left(q-p\right)+c^2\left(r-p\right)\left(r-q\right)=4{\mathrm{\Delta }}^2,$

where $\mathrm{\Delta }$ is the area of the triangle and $a,b,c$ are the lengths of the sides of triangle $ABC$ opposite to angles $A,B$ and $C$ respectively.