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

The co-ordinates of three points $A(-4, 0) ; B(2, 1)$ and $C(3, 1)$ determine the vertices of an equilateral trapezium $ABCD$ . The co-ordinates of the vertex $D$ are :

The equation to the sides of a triangle are $x - 3y = 0$, $4x + 3y = 5$ and $3x + y = 0$. The line $3x - 4y = 0$ passes through

If the straight lines $x + 3y = 4,\,\,3x + y = 4$ and $x +y = 0$ form a triangle, then the triangle is

If the locus of the point, whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5: 4$, is $a x^2+b y^2+c x y+d x+e y+170=0$, then the value of $\mathrm{a}^2+2 \mathrm{~b}+3 \mathrm{c}+4 \mathrm{~d}+\mathrm{e}$ is equal to:

Let the equation of two sides of a triangle be $3x\,-\,2y\,+\,6\,=\,0$ and $4x\,+\,5y\,-\,20\,=\,0.$ If the orthocentre of this triangle is at $(1, 1),$ then the equation of its third side is