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Question

For the straight lines 4x+3y6=0 and 5x+12y+9=0, find the Bisector of the angle which contains (0,0)

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Solution

Equations of bisectors of the angles between the given lines are
4x+3y642+32=±5x+12y+952+122
4x+3y616+9=±5x+12y+925+144
4x+3y625=±5x+12y+9169
4x+3y65=±5x+12y+913
52x+39y78=±(25x+60y+45)
52x+39y78=25x+60y+45,52x+39y78=25x60y45
27x+21y123=0 and 77x+99y33=0
9x+7y41=0 and 7x+9y3=0
If θ is the acute angle between the line 4x+3y6=0 and the bisector 9x7y41=0, then
tanθ=∣ ∣ ∣ ∣43971+(43)97∣ ∣ ∣ ∣=28272136=5515=113>1
Hence the bisector of the angle containing the origin is
4x3y+6(4)2+(3)2=5x+12y+952+122
or 4x3y+625=5x+12y+9169
or 4x3y+65=5x+12y+913
or 52x39y+78=25x+60y+45
or 77x99y+33=0
or 7x+9y3=0


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