Question 1086403
{{{graph(300,300,-10,10,-10,10,(24-4x)/3,(5x+30)/12)}}} shows {{{red(4x+3y-24=0)}}} and {{{green(5x-12y+30=0)}}} lines.
Those lines form four angles.
Because those line are are not perpendicular to each other,
two of those angles are acute and two are obtuse.
 
The bisector of an angle is the locus of the points that are
at equal distance from the two sides of the angle.
Distance from a point {{{P(x[P],y[P])}}} to a line {{{ax+by+c=0}}} is
{{{abs(ax[P]+by[P]+c)/sqrt(a^2+b^2)}}} .
So, the lines bisecting the angles formed by {{{4x+3y-24=0}}} and {{{5x-12y+30=0}}}
have the equations
{{{abs(4x+3y-24)/sqrt(4^2+3^2)}}} {{{"="}}} {{{abs(5x-12y+30)/sqrt(5^2+12^2)}}} .
That simplifies to
{{{abs(4x+3y-24)/5}}} {{{"="}}} {{{abs(5x-12y+30)/13}}} ,
and to
{{{13(4x+3y-24)}}}{{{"="}}}{{{" " +- 5(5x-12y+30)}}}
and
{{{52x+39y-312}}}{{{"="}}}{{{" " +- (25x-60y+150)}}} .
The plus sign would give us
{{{52x+39y-312=25x-60y+150}}} <--> {{{52x-25x+39y+60y-312-150=0}}} <--> {{{27x+99y-462=0}}} <--> {{{9x+33y-154=0}}} ,
with a slope of {{{-9/33}}} {{{"="}}} {{{-3/11}}} ,
showing a slight downward slope.
The minus sign would give us
{{{52x+39y-312=-25x+60y-150}}} <--> {{{52x+25x+39y-60y-312+150=0}}} <--> {{{77x-21y-162=0}}} ,
with a slope of {{{77/21}}} {{{"="}}} {{{11/3}}} ,
showing a very steep upwards slope.
 
The line {{{green(5x-12y+30=0)}}} has a shallow upwards slope of {{{green(5/12)}}} .
The line {{{red(4x+3y-24=0)}}} has a slope of {{{red(-4/3)}}} (downwards).
A line perpendicular to {{{4x+3y-24=0}}} would have a steeper upwards slope of
{{{3/4=9/12>5/12}}} .
So, the acute angle between {{{5x-12y+30=0}}} and {{{4x+3y-24=0}}}
includes shallower slopes between {{{5/12}}} and {{{-4/3}}} ,
such as {{{blue(-3/11)}}} , not steeper slopes, such as {{{3/4}}} or {{{11/3}}} .
That means that the acute angle bisector line is
{{{highlight(blue(9x+33y-154=0))}}} 
 
{{{graph(300,300,-10,10,-10,10,(24-4x)/3,(5x+30)/12,(154-9x)/33)}}}