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:

Let $A \equiv (0, 1)$,$B \equiv (2, 0)$,and point $P$ be a point on the line $4x + 3y + 9 = 0$. Find the coordinates of point $P$ such that $|PA - PB|$ is maximized.

$A$ 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 equation of the line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes is given by

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:

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1, -2)$,then the equation of line $BC$ is