If $(1, 5)$ is the midpoint of the segment of a line between the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$, then the equation of the line is:

Two non-parallel sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x-5y=6$,then one of the possible values of $(\alpha+\beta)$ is

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

The point on the line $3x + y + 4 = 0$ which is equidistant from $(-5, 6)$ and $(3, 2)$ is

$A$ straight line $L$ passes through the point of intersection of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. If $L$ intersects the positive $x$-axis at $A(a, 0)$ and the positive $y$-axis at $B(0, b)$, then the minimum area of the triangle $OAB$ (where $O$ is the origin) is ....

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines,$x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$ and $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$ respectively,then $k^{2}$ is equal to :