The orthocentre of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ is

The coordinates of three points $A(-4, 0)$,$B(2, 1)$,and $C(3, 1)$ determine the vertices of an isosceles trapezium $ABCD$. The coordinates of the vertex $D$ are:

The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,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.

If the orthocentre of the triangle formed by the lines $y = x + 1$,$y = 4x - 8$,and $y = mx + c$ is at $(3, -1)$,then $m - c$ is :

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: