The sides of a rhombus $ABCD$ are parallel to the lines, $x - y + 2\, = 0$ and $7x - y + 3\, = 0$. If the diagonals of the rhombus intersect at $P( 1, 2)$ and the vertex $A$ ( different from the origin) is on the $y$ axis, then the ordinate of $A$ is

The opposite angular points of a square are $(3,\;4)$ and $(1,\; - \;1)$. Then the co-ordinates of other two points are

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side 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 $\alpha, \beta, \gamma, \delta \in \mathrm{Z}$ and let $\mathrm{A}(\alpha, \beta), \mathrm{B}(1,0), \mathrm{C}(\gamma, \delta)$ and $D(1,2)$ be the vertices of a parallelogram $\mathrm{ABCD}$. If $\mathrm{AB}=\sqrt{10}$ and the points $\mathrm{A}$ and $\mathrm{C}$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to

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