The point $P(3,6)$ is first reflected on the line $y=x$ and then the image point $Q$ is again reflected on the line $y=-x$ to get the image point $Q^{\prime}$. Then,the circumcentre of the $\Delta P Q Q^{\prime}$ is

Let $A(1, 1)$,$B(-4, 3)$,and $C(-2, -5)$ be the vertices of a triangle $ABC$. Let $P$ be a point on the side $BC$,and let $\Delta_{1}$ and $\Delta_{2}$ be the areas of triangle $APB$ and triangle $ABC$,respectively. If $\Delta_{1} : \Delta_{2} = 4 : 7$,then find the area enclosed by the lines $AP$,$AC$,and the $x$-axis.

$A$ straight line $4x+y-1=0$ passing through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. Then the equation of the line $AC$ such that $AB=AC$,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 $\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4}=$

$\Delta ABC$ has integral side lengths. The internal angle bisector of angle $B$ meets side $AC$ at $D$. If $AD = 3$ and $DC = 8$,then the perimeter of $\Delta ABC$ is:

Suppose $\triangle ABC$ is an isosceles triangle with $\angle C=90^{\circ}$,$A=(2,3)$ and $B=(4,5)$. Then the centroid of the triangle is