Find the length of the perpendicular drawn from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$.

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point on the line $7x - 9y + 10 = 0$ to the lines $3x + 4y = 5$ and $12x + 5y = 7$,respectively. Then,

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

If the points $(1, 2)$ and $(3, 4)$ lie on opposite sides of the line $3x - 5y + a = 0$,then:

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$,where $k \neq 0$,then the distance from the origin to the line $l$ is