Question 1011464
The lines look like this:
{{{drawing(400,400,-1,11,-1,11,
grid(0),
blue(line(-1,12,5,-12)),
green(line(-1,3.333,11,-0.6667)),
locate(1,10,blue(4x+y=8)),
locate(1,9,green(x+3y=9)),
red(line(-1,5.098,11,-7.86))
)}}}
For each line, all 3 coefficients are positive,
so the slope is negative and part of the line is in the first quadrant.
As we see in the drawing, the same will be true for the bisector of the acute angles.
If {{{P(x,y)}}} is a point in the angle bisector,
{{{P}}} is at equal distance from the lines forming the angle.
The distance from a point {{{P(x,y)}}} to a line {{{ax+by+c=0}}} id
{{{abs(ax+by+c)/sqrt(a^2+b^2)}}} ,
so for a point in the bisector of an angle formed by lines
{{{x+3y=9}}}<-->{{{x+3y-9=0}}} and
{{{4x+y=8}}}<-->{{{4x+y-8=0}}} ,
{{{abs(x+3y-9)/sqrt(1^2+3^2)=abs(4x+y-8)/sqrt(4^2+1^2)}}}
{{{abs(x+3y-9)/sqrt(10)=abs(4x+y-8)/sqrt(17)}}}
{{{sqrt(17)*abs(x+3y-9)=sqrt(10)*abs(4x+y-8)}}} .
Two lines that are not parallel form, two angles, each with its own bisector.
In this case the bisectors are
{{{sqrt(17)(x+3y-9)=sqrt(10)*(4x+y-8)}}}<-->{{{(sqrt(17)-4*sqrt(10))x+(3sqrt(17)-sqrt(10))y=9sqrt(17)-8sqrt(10)}}} and
{{{sqrt(17)(x+3y-9)=-sqrt(10)*(4x+y-8)}}}<-->{{{highlight((sqrt(17)+4*sqrt(10))x+(3sqrt(17)+sqrt(10))y=9sqrt(17)+8sqrt(10))}}}
The one with all positive coefficients is the bisector of the acute angles.
The bisector of the obtuse angles is perpendicular and has a positive slope (coefficients of {{{x}}} an {{{y}}} have opposite signs).