The locus of the mid-point of the segment intercepted between the axes by the variable line $x \cos \alpha + y \sin \alpha = p$,where $p$ is a constant,is

$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ 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:

$ABC$ is a variable triangle such that $A$ is $(1, 2)$,and $B$ and $C$ lie on the line $y = x + \lambda$ (where $\lambda$ is a variable). The locus of the orthocenter of triangle $ABC$ is:

$A$ point on the straight line $3x + 5y = 15$ which is equidistant from the coordinate axes will lie only in