The vertices of `DeltaABC` are `A(-2,4)`, `B(5,5)` and `C(4,-2)`. Find the equation of the bisector of `/_A`.

Explanation

`AB=sqrt((-2-5)^(2)+(4-5)^(2))`
`=sqrt(49+1)=5sqrt(2)`
and `AC=sqrt((-2-4)^(2)+(4+2)^(2))`
`=sqrt(36+36)=6sqrt(2)`

The bisector line of `/_A` meets the line `BC` at `D`
and `(BD)/(DC)=(AB)/(AC)=(5sqrt(2))/(6sqrt(2))=5 : 6`
Divide `BC` in the ratio `5 : 6` and thus the co-ordinates of `D` are
`((5xx4+6xx5)/(5+6),(5xx(-2)+6xx5)/(5+6))=((50)/(11),(20)/(11))`
Now, equation of `AD` is
`y-4=((20)/(11)-4)/((50)/(11)-(-2))(x+2)`
`=(-24)/(72)(x+2)`
`=-(1)/(3)(x+-2)`
`implies 3y-12=-x-2`
`implies x+3y=10`
Which is the required equation.