According to angle bisector theorem in a triangle ABC,

The bisector of angle A will meet at BC say at a point called X

And hence, AX divides BC in the ratio AC:AB

So, according to the theorem AC:AB=CX:XB

Lets find AC:AB b y distance formula,

AC= √(x2-x1)2 +(y2-y1)2 = √(2-4)2 +(3-3)2= 2

Now find AB :

AB= √(x2-x1)2 +(y2-y1)2= √(4-0)2 + (3-0)2= 5

Simply from (1) and (2) we get:

AC:AB=CX:XB

=> 2:5=CX:CB

Now according to section formula:

Let X be (x,y)

Let the ratio in which BC is divided by X be m:n

So m= 2

n = 5

x1,y1=2,3

x2,y2=0,0

x= 10/7 & y= 15/7

X=(10/7),(15/7)

Now, calculate the slope of AX:

x1,y1=4,3

x2,y2=(10/7,15/7)

Slope, m= y2-y1/x2-x1 = 1/3

Now, this is really tiring:

y-y1=m(x-x1)

Now A = (4,3)

==> x1=4

==>y1=3

y-3=1/3(x-4)

==> 3(y-3)=x-4

==>3y-9=x-4

==>3y-x-5=0

That's the equation:

x-3y+5=0