$A(-2, 3)$ is a point on the line $4x + 3y - 1 = 0$. If the points on the line that are $10$ units away from the point $A$ are $(x_1, y_1)$ and $(x_2, y_2)$,then $(x_1 + y_1)^2 + (x_2 + y_2)^2 =$

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and the $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the line $AB$ at $C, D, E$ respectively. If $O$ is the origin,then the area of the quadrilateral $OCEB$ is:

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is

Let the lines $3x - 4y - \alpha = 0$,$8x - 11y - 33 = 0$,and $2x - 3y + \lambda = 0$ be concurrent. If the image of the point $(1, 2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\frac{57}{13}, \frac{-40}{13}\right)$,then $|\alpha \lambda|$ is equal to:

Suppose $P$ and $Q$ lie on $3x + 4y - 4 = 0$ and $5x - y - 4 = 0$ respectively. If the mid-point of $PQ$ is $(1, 5)$,then the slope of the line passing through $P$ and $Q$ is

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.