Area of the parallelogram whose sides are $x\cos \alpha + y\sin \alpha = p$ $x\cos \alpha + y\sin \alpha = q,\,\,$ $x\cos \beta + y\sin \beta = r$ and $x\cos \beta + y\sin \beta = s$ is

The points $(1, 3)$ and $(5, 1)$ are the opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c,$ then the value of c will be

The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :

If the equation of the locus of a point equidistant from the points $({a_1},{b_1})$ and $({a_2},{b_2})$ is $({a_1} - {a_2})x + ({b_1} - {b_2})y + c = 0$, then the value of $‘c’$ is

The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. unit. The locus of the point is