Question 883651
You are only given slopes, so it must mean that y-intercepts do not matter.
The lines could intersect each other anywhere, and they could intersect the x- and y-axes anywhere,
but for convenience I will make the lines intersect at the origin.
(Whoever does not like my drawing can move the axes up, down, and left or right to his/her taste. As long as nothing is rotated, the slopes remain the same).
{{{drawing(300,300,-2,3,-1,4,
grid(0),triangle(0,0,1,0,1,2),
rectangle(1,0,0.9,0.1),locate(1.02,1.2,2),
line(-1,-2,2,4),line(0,-1,0,4),
locate(1.9,2.5,line),locate(2.5,2.5,1),
arrow(1.8,2.36,1.2,2.36),
locate(-2,2.8,line),locate(-1.3,2.8,2),
arrow(-1,2.66,-0.1,2.66),
green(line(-0.3,-1.27,1,4.236)),
locate(1.9,3.5,green(bisector)),
green(arrow(1.85,3.36,0.8,3.36)),
red(arc(0,0,1.5,1.5,-63.4,0)),
locate(0.5,0.4,red(A))
)}}} The slope of line 1 is the tangent of the angle {{{red(A))}}} it makes with the x-axis (measured counterclockwise from the positive x-axis).
The slope of the y-axis is undefined, and line 2 is the y-axis.
The bisector of the acute angle formed by lines 1 and 2 is drawn in green.
The angle it forms with the positive x-axis is
{{{(A+90^o)/2}}}
(If you don't like degrees, it is {{{(A+pi/2)/2}}} in radians).
The slope of the bisector is the tangent of that angle.
 
If you only need an approximate value, you could use inverse trigonometric functions.
For an exact value, I would use trigonometric identities.
 
USING INVERSE FUNCTIONS:
We use the inverse of tangent from calculator or computer to find the angle that has a tangent of {{{2}}}.
Hopefully you know how to do that with your calculator and/or computer.
The inverse function of tangent is represented as {{{"tan"^(-1)}}} , or {{{arctan}}} , or {{{ATAN}}} , and in some calculators you have to press a key for inverse functions and then the tangent key.
{{{tan(A)=2}}}--->{{{A=63.4^o}}}(rounded),
or {{{A=1.107}}}(rounded) in radians, if you prefer radians.
Using {{{A=63.4^o}}}:
{{{(A+90^o)/2=76.7^o}}} and {{{tan(76.7^o)=4.23}}}(rounded) is the approximate value for the slope of the bisector.
Using {{{A=1.107}}}(in radians):
{{{(1.107+pi/2)/2=1.339}}}(rounding the result but not {{{pi}}} ), and
{{{tan(1.339)=4.24}}}(rounded) is the approximate value for the slope of the bisector.
You could round to more digits to get a better approximation.
 
USING TRIGONOMETRIC IDENTITIES FOR EXACT VALUE:
From the values of the trigonometric functions for {{{A}}},
and the well known values for {{{90^o=pi.2}}} ,
we would like to find the trigonometric functions for {{{(A+90^o)/2=(A+pi/2)/2}}} .
For easier typing, I am going to use {{{90^o rather {{{pi/2}}} from now on.
There are probably many ways to get to {{{tan((A+90^o)/2)}}} ,
but I am going to show the one that came to my mind first,
and that requires starting from {{{sin(A)}}} and {{{cos(A)}}} .
The right triangle with angle {{{A}}} that I included in my sketch
has legs measuring {{{1}}} and {{{2}}} ,
so the Pythagorean theorem says that the length of the hypotenuse is {{{sqrt(1^2+2^2)=sqrt(1+4)=sqrt(5)}}} .
Trigonometric ratios tell us that
{{{sin(A)=2/sqrt(5)}}} and {{{cos(A)=1/sqrt(5)}}}
Trigonometric identities for sum of angles tell us that
{{{cos(A+90^o)=cos(A)cos(90^o)-sin(A)sin(90^o)=(1/sqrt(5))*0-(2/sqrt(5))*1=-2/sqrt(5)}}}
We are half of he way there
Trigonometric identities for half angles are
{{{abs(sin(B/2))=sqrt((1-cos(B))/2)}}} and {{{abs(cos(B/2))=sqrt((1+cos(B))/2)}}} ,
and we have to figure out the sign for ourselves.
Since {{{(A+90^o)/2}}} is in the first quadrant, we know that all its trigonometric functions have positive values, so
{{{sin((A+90^o)/2)=sqrt((1-cos(A+90^o))/2)=sqrt((1-(1/sqrt(5)))/2)}}}
{{{cos((A+90^o)/2)=sqrt((1+cos(A+90^o))/2)=sqrt((1+(1/sqrt(5)))/2)}}}
Tangent is sine divided by cosine, so
{{{tan((A+90^o)/2)=sqrt((1-(2/sqrt(5)))/2)/sqrt((1+(2/sqrt(5)))/2)=sqrt((1-(2/sqrt(5)))/(1+(2/sqrt(5))))}}}={{{sqrt((sqrt(5)+2)/(sqrt(5)-2))=sqrt((sqrt(5)+2)(sqrt(5)+2)/((sqrt(5)-2)(sqrt(5)+2)))}}}={{{sqrt((sqrt(5)+2)^2/((sqrt(5)+2)(sqrt(5)-2)))}}}={{{sqrt((sqrt(5)+2)^2/(5-2^2))}}}={{{sqrt((sqrt(5)+2)^2/(5-4))}}}={{{sqrt((sqrt(5)+2)^2)}}}={{{highlight(sqrt(5)+2)}}}
The approximate value of {{{sqrt(5)+2}}} is {{{4.236}}} .