Find the equation of the bisector of the acute angle between the lines, 5x + 12y = 0 and 3x - Maths - Straight Lines

Question

Find the equation of the bisector of the acute angle between the
lines, 5x + 12y = 0 and 3x = 4y
I tried solving this problem but a1a2 +
b1b2 was equal to 0 Then in this case which
equation would be taken as the equation of acute angle bisector
between the two lines?

Akshay Kamthan · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

Equation of bisector of lines a1x+b1y+c1=0 and a2x+b2y+c2=0 is
given by:a1x+b1y+c1a12+b12=±a2x+b2y+c2a22+b22 ;iNow if
a1a2+b1b2>0, then negative sign in i will give acute angled
bisectorand if a1a2+b1b2<0, then positive sign in i will give
acute angled bisector We have3x-4y=05x+12y=0
a1a2+b1b2=15-48=-33<0, hence positive sign will give acute
angled bisector Hence equation of acute angled bisector
is:3x-4y32+42=-5x+12y122+52⇒3x-4y9+16=5x+12y144+25⇒3x-4y5=5x+12y13⇒39x-52y=25x+60y=14x=112yDividing
each term by 14 we get
x=8ypls check a1a2+b1b2 is not zero

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Find the equation of the bisector of the acute angle between the lines, 5x + 12y = 0 and 3x = 4y. I tried solving this problem but a1a2 + b1b2 was equal to 0. Then in this case which equation would be taken as the equation of acute angle bisector between the two lines?

Find the equation of the bisector of the acute angle between the lines, 5x + 12y = 0 and 3x = 4y.

I tried solving this problem but a₁a₂ + b₁b₂ was equal to 0. Then in this case which equation would be taken as the equation of acute angle bisector between the two lines?