If the vertices of a triangle $ABC$ are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$,then the equation of the internal angle bisector of $\angle ABC$ is

If $(0, 1), (1, 1)$ and $(1, 0)$ are the midpoints of the sides of a triangle,find its incentre.

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

The incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ 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:

In a $\triangle ABC$,$2x+3y+1=0$ and $x+2y-2=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively. If $A=(3,2)$,then the equation of the side $BC$ is