SOLUTION: what is the equation of the line bisector of the acute angle formed by the intersection of the lines 4x+3y-24=0 and 5x-12y+30=0

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Question 1086403: what is the equation of the line bisector of the acute angle formed by the intersection of the lines 4x+3y-24=0 and 5x-12y+30=0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%2824-4x%29%2F3%2C%285x%2B30%29%2F12%29 shows red%284x%2B3y-24=0%29 and green%285x-12y%2B30=0%29 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%28x%5BP%5D%2Cy%5BP%5D%29 to a line ax%2Bby%2Bc=0 is
abs%28ax%5BP%5D%2Bby%5BP%5D%2Bc%29%2Fsqrt%28a%5E2%2Bb%5E2%29 .
So, the lines bisecting the angles formed by 4x%2B3y-24=0 and 5x-12y%2B30=0
have the equations
abs%284x%2B3y-24%29%2Fsqrt%284%5E2%2B3%5E2%29 %22=%22 abs%285x-12y%2B30%29%2Fsqrt%285%5E2%2B12%5E2%29 .
That simplifies to
abs%284x%2B3y-24%29%2F5 %22=%22 abs%285x-12y%2B30%29%2F13 ,
and to
13%284x%2B3y-24%29%22=%22%22+%22+%2B-+5%285x-12y%2B30%29
and
52x%2B39y-312%22=%22%22+%22+%2B-+%2825x-60y%2B150%29 .
The plus sign would give us
52x%2B39y-312=25x-60y%2B150 <--> 52x-25x%2B39y%2B60y-312-150=0 <--> 27x%2B99y-462=0 <--> 9x%2B33y-154=0 ,
with a slope of -9%2F33 %22=%22 -3%2F11 ,
showing a slight downward slope.
The minus sign would give us
52x%2B39y-312=-25x%2B60y-150 <--> 52x%2B25x%2B39y-60y-312%2B150=0 <--> 77x-21y-162=0 ,
with a slope of 77%2F21 %22=%22 11%2F3 ,
showing a very steep upwards slope.

The line green%285x-12y%2B30=0%29 has a shallow upwards slope of green%285%2F12%29 .
The line red%284x%2B3y-24=0%29 has a slope of red%28-4%2F3%29 (downwards).
A line perpendicular to 4x%2B3y-24=0 would have a steeper upwards slope of
3%2F4=9%2F12%3E5%2F12 .
So, the acute angle between 5x-12y%2B30=0 and 4x%2B3y-24=0
includes shallower slopes between 5%2F12 and -4%2F3 ,
such as blue%28-3%2F11%29 , not steeper slopes, such as 3%2F4 or 11%2F3 .
That means that the acute angle bisector line is
highlight%28blue%289x%2B33y-154=0%29%29