The area of triangle formed by the lines $x = 0,y = 0$ and $\frac{x}{a} + \frac{y}{b} = 1$, is

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid - point of $BC$ is $(5, 6)$ then the equation of $BC$ is :

The number of possible straight lines , passing through $(2, 3)$ and forming a triangle with coordinate axes, whose area is $12 \,sq$. units , 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:

Let the equations of two adjacent sides of a parallelogram $A B C D$ be $2 x-3 y=-23$ and $5 x+4 y$ $=23$. If the equation of its one diagonal $AC$ is $3 x +$ $7 y=23$ and the distance of A from the other diagonal is $d$, then $50 d ^2$ is equal to $........$.