In the figure given below,$ABCDEF$ is a regular hexagon of side length $1$,$AFPS$ and $ABQR$ are squares. Then,the ratio $\frac{\operatorname{ar}(APQ)}{\operatorname{ar}(SRP)}$ equals

The points $A(2, 1)$,$B(3, -2)$ and $C(a, b)$ are vertices of the rectangle $ABCD$. If the point $P(3, 4)$ lies on the line $CD$,then $5a + 10b = $

Consider a $\triangle ABC$ in the $XY$-plane with vertices $A=(0,0)$,$B=(1,1)$,and $C=(9,1)$. If the line $x=a$ divides the triangle into two parts of equal area,then $a$ equals

Let $ABC$ be a triangle such that the equations of lines $AB$ and $AC$ are $3y-x=2$ and $x+y=2$,respectively,and the points $B$ and $C$ lie on the $x$-axis. If $P$ is the orthocentre of the triangle $ABC$,then the area of the triangle $PBC$ is equal to

Suppose the vertices of a triangle are given by $A(0,3)$,$B(-2,0)$,and $C(6,1)$. For the point $(\alpha, \alpha+1)$ to lie inside the triangle,$\alpha$ must lie in the interval:

The area of the triangle formed by the lines $y = m_1x + c_1$,$y = m_2x + c_2$ and $x = 0$ is: