The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$, 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

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is

A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y -$ axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle OAB$ is $\frac{98}{3} \sqrt{3}$, then $a ^2- b ^2$ is equal to:

Find the area of the triangle formed by the line $y-x=0, x+y=0$ and $x-k=0$.