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 -

One vertex of the equilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is

The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 \mathrm{x}+\mathrm{y}=14$ and $3 \mathrm{x}-2 \mathrm{y}=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. The equation of the side $\mathrm{BC}$ is :

The area of triangle formed by the lines $x + y - 3 = 0 , x - 3y + 9 = 0$ and $3x - 2y + 1= 0$

A vertex of equilateral triangle is $(2, 3)$ and equation of opposite side is $x + y = 2,$ then the equation of one side from rest two, is

If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex