Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

$A$ straight line passes through a fixed point $(h, k)$. The locus of the foot of the perpendicular drawn from the origin to this line is:

$A$ 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

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

Let $A$ be a fixed point $(0,6)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $AB$ and the perpendicular bisector of $AB$ meets the $y$-axis at $C$. The locus of the mid-point $P$ of $MC$ is:

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: