The angle between the lines 5x+3y+5=0 and 4x-6y+7=0 is
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Answer:
Equations of bisectors of the angles between the given lines are
4
2
+3
2
4x+3y−6
=±
5
2
+12
2
5x+12y+9
⇒
16+9
4x+3y−6
=±
25+144
5x+12y+9
⇒
25
4x+3y−6
=±
169
5x+12y+9
⇒
5
4x+3y−6
=±
13
5x+12y+9
⇒52x+39y−78=±(25x+60y+45)
⇒52x+39y−78=25x+60y+45,52x+39y−78=−25x−60y−45
⇒27x+21y−123=0 and 77x+99y−33=0
⇒9x+7y−41=0 and 7x+9y−3=0
If θ is the acute angle between the line 4x+3y–6=0 and the bisector 9x–7y–41=0, then
tanθ=
∣
∣
∣
∣
∣
∣
∣
∣
∣
1+(−
3
4
)
7
9
3
−4
−
7
9
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
21−36
−28−27
∣
∣
∣
∣
∣
=
15
55
=
3
11
>1
Hence the bisector of the acute angle is 7x+9y−3=0
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