Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

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

If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is

Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positive $\mathrm{x}$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is :

Let $A (-3, 2)$ and $B (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$, then the vertex $C$ lies on the line