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Question

Find the angle between each of the following pairs of straight lines:

(i) 3x+y+12=0 and x+2y1=0

(ii) 3xy+5=0 and x3y+1=0

(iii) 3x+4y7=0 and 4x3y+5=0

(iv) x4y=3 and 6xy=11

(v) (m2mn)y=(mn+n2)x+n2 and (mn+m2)y=(mn+n2)x+m3.


Solution

(i) The given equations are

3x+y+12=0

x+2y1=0

Writing the equation in the form

y=mx+c

3x+y+12=0

y=3x12

m1=3

Also,

x+2y1=0

2y=1x

y=12x2

m2=12

Angle between the lines

tan θ=m1m21+m1m2

=∣ ∣3(12)1+(3)(12)∣ ∣

=3+121+32=6+1222+32

=55=1

angle=π4=45

(ii) Finding slopes of the lines by converting the equation in the form

y=mx+c

3xy+5=0

y=3x+5

m1=3

Also,

x3y+1=0

3y=x+1

y=x3+13

m2=13

Thus angle between the lines is

tan θ=m1m21+m1m2

=3131+313=9131+1

=832=86=43

θ=tan1(43)

(iii) 3x+4y7=0 and 4x3y+5=0

To find angle between the lines, convert the equations in the form

y=mx+c

3x+4y7=0

4y=3x+7

y=34x+74

m1=34

Also, 4x3y+5=0

3y=4x+5

y=43x+53

m1=43

The angle between the line is given by

tan θ=m1m21+m1m2

=∣ ∣34431+(3)4(43)∣ ∣=344311

θ=π2 or 90

(iv) x4y=3 and 6xy=11

To find angle convert the equation in the form y=mx+c

x4y=3

4y=x3

y=x434

m1=14

Also, 6xy=11

y=6x11

m2=6

Thus, angle between the lines is

tan θ=m1m21+m1m2

=1461+14×6

=2341+32=23452

θ=tan1 (2310)

(v) (m2mn)y=(mn+n)2x+n2 and (nm+m2)y=(mn+n2)x+m3

Converting the equation in the form

y=mx+c

y=(mn+n2)m2mnx+n3(m2mn)

m1=mn+n2m2mn

Also, y=(mnn2)nm+m2x+m3nm+m2

m2=mnn2nm+m2

Thus, angle between 2 lines is tan θ

tan θ=m1m21+m1m2

=∣ ∣(mn+n2m2mn)(mnn2nm+m2)1+(mn+n2m2mn)(mnn2nm+m2)∣ ∣

=m2n2+m3n+n3m+n2m2m3n+m2n2+n2m2mn3m3n+m4m2n2m3n+m2n2nm3+mn3n4

=4m2n2m4n4

θ=tan14m2n2m4n4

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