The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ 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.

Find the area of the triangle formed by the lines $y-x=0$,$x+y=0$,and $x-k=0$.

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

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are

Let the mirror image of point $A(\alpha, \beta)$ in the line mirror $x + 2y = 3$ be point $B$,and the image of $B$ in the line $3x - 2y = 5$ be $C$. If the origin is the orthocentre of triangle $ABC$ and $P(a, b)$ is a point inside the triangle such that triangles $PAB$,$PBC$,and $PCA$ have the same area,then $3(a + b)$ is: