The angular points of a triangle are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$. The equation of the bisector of the angle $\angle ABC$ 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.

Let the circumcentre of a triangle with vertices $A(a, 3)$,$B(b, 5)$,and $C(a, b)$,where $ab > 0$,be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_{1}, k_{2})$,then $k_{1} + k_{2}$ is equal to.

The points $\left( \frac{a}{\sqrt{3}}, a \right)$,$\left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $\left( \frac{a}{\sqrt{3}}, 3a \right)$ are the vertices of:

Two lines are drawn through $(3, 4)$,each of which makes an angle of $45^\circ$ with the line $x - y = 2$. The area of the triangle formed by these lines is:

The area of the triangle formed by the lines $x = 0$,$y = 0$,and $x/a + y/b = 1$ is: