The area of the parallelogram formed by the lines $y = mx$,$y = mx + 1$,$y = nx$,and $y = nx + 1$ is equal to

For $a > b > c > 0$,the distance between $(1,1)$ and the point of intersection of the lines $ax + by + c = 0$ and $bx + ay + c = 0$ is less than $2\sqrt{2}$. Then:

Let $P$ be the point of intersection of the lines $L_1 \equiv x-y-7=0$ and $L_2 \equiv x+y-5=0$. $A(x_1, y_1)$ and $B(x_2, y_2)$ are points on the lines $L_1=0$ and $L_2=0$ respectively such that $PA=3\sqrt{2}$,$PB=\sqrt{2}$,$x_1, y_1 \geq 0$,$x_2, y_2 \geq 0$. Then the angle made by the line segment $AB$ at the origin is

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

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

$A$ straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance $1$ unit from the origin intersects the straight line $y=2x+3+\sqrt{2}$ at $(x_0, y_0)$. Then $2x_0+y_0$ is equal to