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
$x + y = 1$
$x + y = 2$
$x + y = 2xy$
$2x + 2y = 1$
The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is
The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -
The area enclosed within the curve $|x| + |y| = 1$ is
If the straight line $ax + by + c = 0$ always passes through $(1, -2),$ then $a, b, c$ are
If the middle points of the sides $BC,\, CA$ and $AB$ of the triangle $ABC$ be $(1, 3), \,(5, 7)$ and $(-5, 7)$, then the equation of the side $AB$ is