Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .... .

The centre of a square of side $4$ units length is $(3,7)$ and one of the diagonals is parallel to the line $y=x$. If $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ are the vertices of this square,then $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $(3,4)$. If $BD=2 \sqrt{2}$,$A=(1,2)$,$B=(\alpha, \beta)$,$D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$,then $\beta+\gamma-\delta=$

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a

Let $ABCD$ be a quadrilateral such that there exists a point $E$ inside the quadrilateral satisfying $AE=BE=CE=DE$. Suppose $\angle DAB, \angle ABC, \angle BCD$ are in an arithmetic progression. Then the median of the set $\{\angle DAB, \angle ABC, \angle BCD\}$ is