The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$.

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

Given the family of lines,$a(2x + y + 4) + b(x - 2y - 3) = 0$. Among the lines of the family,the number of lines situated at a distance of $\sqrt{10}$ from the point $M(2, -3)$ is:

$A(-2, 3)$ 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 number of straight lines that can be drawn through the point $(2, 5)$ which form a triangle of area $24 \text{ sq. units}$ with the coordinate axes is:

Let $A(-1, 1)$,$B(3, 4)$,and $C(2, 0)$ be three given points. $A$ line $y = mx$,$m > 0$,intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively,such that $A_1 = 3A_2$. Then the value of $m$ is equal to: