`A(1,3)and c(-2/5,-2/5)`are the vertices of a `DeltaABCand`the equation of the angle bisector of `/_ABC` is `x+y=2.`

A(1,3) and c(-(2)/(5),-(2)/(5)) are the vertices of a Delta ABC and the equation of the angle bisector of /_ABC is x+y=2 .

A (1,3) and C (-2/5,-2/5) are the vertices of a triangle ABC and the equation of the angle bisector of angleABC is x+y =2 Equation of side AB is

A (1,3) and C (-2/5,-2/5) are the vertices of a triangle ABC and the equation of the internal angle bisector of angleABC is x+y =2 Equation of BC is

A(1,3) and C(-2,/5, -2/5) are the vertices of a triangle ABC and the equation of the internal angle bisector of angleABC " is " x+y=2. The equation of side BC is

A(1,3) and C(-2/5, -2/5) are the vertices of a triangle ABC and the equation of the internal angle bisector of angleABC " is " x+y=2. The coordinates of vertex B are

A (1,3) and C (-2/5,-2/5) are the vertices of a triangle ABC and the equation of the internal angular bisector of angleABC is x+y -2=0 Coordinates of vertex B

If A (1,-2), B (-2,3) and C(2,-5) are the vertices of Delta ABC, then the equation of the median BE is

A(5, 2), B(2, 3) and C(6, 5) are the verticies of Delta ABC . Find equation of internal bisector of angle BAC .

If A(4, 3), B(0, 0) and C(2, 3) are the vertices of a triangle ABC , find the equation of the bisector of angle A .

If A(1, -1, -3), B(2, 1, -2), C(-5, 2, -6) are the vertices of a DeltaABC , then the length of internal bisector of angle A is