Find the equation of the line which is equidistant from the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.

Let the line $L$ drawn perpendicular to the lines $2x - 3y + 4 = 0$ and $6x - 9y + 7 = 0$ meet them at $A$ and $B$ respectively. If $P(1, 1)$ is a point on $L$,then the ratio in which $P$ divides $AB$ is

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then $\frac{1}{a^2}+\frac{1}{b^2}=$

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

The length of the perpendicular from the point $P(a, b)$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ is

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is: