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

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.

What type of triangle is formed by the vertices $(-2, 2)$,$(8, -2)$,and $(-4, -3)$?

$\Delta ABC$ 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:

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

The equations of two altitudes of an equilateral triangle are $\sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $\sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The equation of the third altitude is