If $A$ is $(2, 5)$, $B$ is $(4, -11)$ and $ C$ lies on $9x + 7y + 4 = 0$, then the locus of the centroid of the $\Delta ABC$ is a straight line parallel to the straight line is

In a triangle $ABC$, coordianates of $A$ are $(1, 2)$ and the equations of the medians through $B$ and $C$ are $x + y = 5$ and $x = 4$ respectively. Then area of $\Delta ABC$ (in sq. units) is

Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is

The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is 

An equilateral triangle has each of its sides of length $6\,\, cm$ . If $(x_1, y_1) ; (x_2, y_2) \,\, and \,\, (x_3, y_3)$ are its vertices then the value of the determinant,${{\left| {\,\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}\,} \right|}^2}$ is equal to :

The area of triangle formed by the lines $x = 0,y = 0$ and $\frac{x}{a} + \frac{y}{b} = 1$, is