Let a triangle be bounded by the lines $L _{1}: 2 x +5 y =10$; $L _{2}:-4 x +3 y =12$ and the line $L _{3}$, which passes through the point $P (2,3)$, intersect $L _{2}$ at $A$ and $L _{1}$ at $B$. If the point $P$ divides the line-segment $A B$, internally in the ratio $1: 3$, then the area of the triangle is equal to
$\frac{110}{13}$
$\frac{132}{13}$
$\frac{142}{13}$
$\frac{151}{13}$
Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2) .$ If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\Delta \mathrm{PAB}=5\; \mathrm{sq}$ units and it lies on the line, $3 x+y-4 \lambda=0$ then a value of $\lambda$ is
The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ 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 -
The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px+qy= 1$ and $qx + py = 1.$ Then the equation to the median through $A$ is
If $A$ and $B$ are two points on the line $3x + 4y + 15 = 0$ such that $OA = OB = 9$ units, then the area of the triangle $OAB$ is
Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0.$ If the equation to one diagonal is $11x + 7y = 9,$ then the equation of the other diagonal is