Triangle formed by the lines $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ and $24x -7y = 3$ is a/an

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 :

A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is

The co-ordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC (AC = BC)$ are $(-2,3)$ and $(2,0)$ respectively. $A$ line parallel to $AB$ and having a $y$ -intercept equal  to $\frac{43}{12}$ passes through $C$, then the co-ordinates of $C$ are :-

The line $3x + 2y = 24$ meets $y$-axis at $A$ and $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, - 1)$ parallel to $x$-axis at $C$. The area of the triangle $ABC$ is ............... $\mathrm{sq. \, units}$

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