Question

The angle between the lines 5x+3y+5=0 and 4x-6y+7=0 is​

Answer 1

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