$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

Suppose a point $P$ moves so that $BP^2 - AP^2 = 121$,where $A$ and $B$ are $(2, 5)$ and $(5, 11)$ respectively. Then the locus of $P$ is a straight line,whose slope is

If the locus of the centroid of the triangle with vertices $A(a, 0)$,$B(a \cos t, a \sin t)$ and $C(b \sin t, -b \cos t)$ ($t$ is a parameter) is $9x^2 + 9y^2 - 6x = 49$,then the area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

Let $A, B$ and $C$ be three points in a plane. The locus of a point $P$ moving such that $PA^2 + PB^2 = 2PC^2$ is a

The locus of a point which moves such that the area of the triangle formed by it with the vertices $(1, 2)$ and $(-2, 5)$ is $8$ sq. units is/are

Let $a \neq 0, b \neq 0, c$ be three real numbers and $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}, \forall p, q \in \mathbb{R}$. If $L\left(\frac{2}{3}, \frac{1}{3}\right) + L\left(\frac{1}{3}, \frac{2}{3}\right) + L(2, 2) = 0$,then the line $ax + by + c = 0$ always passes through the fixed point: