The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is
$(0,0)$
$\left( {\frac{1}{2},\frac{1}{2}} \right)$
$\left( {\frac{1}{3},\frac{1}{3}} \right)$
$\left( {\frac{1}{4},\frac{1}{4}} \right)$
If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is
A vertex of square is $(3, 4)$ and diagonal $x + 2y = 1,$ then the second diagonal which passes through given vertex will be
Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :
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 , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is
Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$, then the equation of locus of mid- point $P$ of $BC$ is