The value of $k$ for which the lines $7x - 8y + 5 = 0$,$3x - 4y + 5 = 0$ and $4x + 5y + k = 0$ are concurrent is given by

$(a, b)$ is the point of concurrency of the lines $x-3y+3=0$,$kx+y+k=0$,and $2x+y-8=0$. If the perpendicular distance from the origin to the line $L \equiv ax-by+2k=0$ is $p$,then the perpendicular distance from the point $(2, 3)$ to $L=0$ is

If $P \equiv \left( \frac{1}{x_p}, p \right), Q = \left( \frac{1}{x_q}, q \right), R = \left( \frac{1}{x_r}, r \right)$ where $x_k \neq 0$ denotes the $k^{th}$ term of an $H.P.$ for $k \in N$,then:

If the three distinct lines $x + 2ay + a = 0$,$x + 3by + b = 0$,and $x + 4ay + a = 0$ are concurrent,then the point $(a, b)$ lies on a

Let $C$ be the centroid of the triangle with vertices $(3, -1), (1, 3),$ and $(2, 4).$ Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$ Then the line passing through the points $C$ and $P$ also passes through the point

The line parallel to the $x$-axis and passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$,where $(a, b) \ne (0, 0)$ is