I have an idea how to do it.

Are you familiar with the formula for two lines (not perpendicular)?

\tan \theta=\frac{m_1-m_2}{1+m_1m_2}

Where,

m_1,m_2

Are the slopes of the two lines.

~~~~~~

Here is an example of what I said.

~~~~~~

The first line you gave was,

3x+y-5=0

The second line you gave was,

x-3y-1=0

Thus,

y=-3x+5

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

Thus, the angle between them is, (note the negative)

tan \theta = \frac{ 1/3+3}{ 1+(1/3)(-3) }

Notice the numerator is zero () because they are perpendicular. Thus, you now the angle between them is 90 degrees.

Thus, the angle of the bisector needs to do is 45 degree between these two lines.

Let the slope be "m" thus, (between the lower line)

\tan 45 = \frac{ m - 1/3 }{1+m/3}

Thus,

1= \frac{ m - 1/3 }{1 +m/3}

Thus,

1+m/3=m-1/3

Multiply through by 3,

3+m=3m-1

4=2m

2=m

Thus, the slope of the angle bisector is 2.

Now we find the point of intersection (because we already have the slope, if we have a point on a line then we can find its equation).

We do that by solving the equation,

y=-3x+5

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

Thus,

-3x+5=(1/3)x-(1/3)

Multiply by 3,

-9x+15=x-1

16=10x

Thus,

x=1.6

y=-3(1.6)+5=.2

We will use the point intercept formula,

The equation of a line (non-vertical) is,

y-y_0=m(x-x_0)

Where, "m" is the slope and, (x_0,y_0) is a point on line.

Thus,

y-.2=2(x-1.6)

y-.2=2x-3.2

y=2x-3