The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{2{c^2}}}$, 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 -
$x^2 + y^2 = c^2$
$x^2 + y^2 = 2c^2$
$x^2+ y^2 = \frac{c^2}{2}$
$x^2 + y^2 = 4c^2$
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
If the sum of the distances of a point from two perpendicular lines in a plane is $1$, then its locus is
Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is
If one vertex of an equilateral triangle of side $'a'$ lies at the origin and the other lies on the line $x - \sqrt{3} y = 0$ then the co-ordinates of the third vertex are :