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Question

Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y − 1 = 0
(ii) 3x − y + 5 = 0 and x − 3y + 1 = 0
(iii) 3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
(iv) x − 4y = 3 and 6x − y = 11
(v) (m2 − mn) y = (mn + n2) x + n3 and (mn + m2) y = (mn − n2) x + m3.

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Solution

(i) The equations of the lines are

3x + y + 12 = 0 ... (1)

x + 2y − 1 = 0 ... (2)

Let m1 and m2 be the slopes of these lines.

m1=-3, m2=-12

Let θ be the angle between the lines.
Then,

tanθ=m1-m21+m1m2 =-3+121+32 =1θ=π4or 45°

Hence, the acute angle between the lines is 45°

(ii) The equations of the lines are

3x − y + 5 = 0 ... (1)

x − 3y + 1 = 0 ... (2)

Let m1 and m2 be the slopes of these lines.

m1=3, m2=13

Let θ be the angle between the lines.
Then,

tanθ=m1-m21+m1m2 =3-131+1 =43θ=tan-143

Hence, the acute angle between the lines is tan-143.

(iii) The equations of the lines are

3x + 4y − 7 = 0 ... (1)

4x − 3y + 5 = 0 ... (2)

Let m1 and m2 be the slopes of these lines.

m1=-34, m2=43

m1m2=-34×43 =-1

Hence, the given lines are perpendicular.
Therefore, the angle between them is 90°.

(iv) The equations of the lines are

x − 4y = 3 ... (1)

6x − y = 11 ... (2)

Let m1 and m2 be the slopes of these lines.

m1=14, m2=6

Let θ be the angle between the lines.
Then,

tan θ=m1-m21+m1m2 =14-61+32 =2310θ=tan-12310

Hence, the acute angle between the lines is tan-12310

(v) The equations of the lines are

(m2 − mn) y = (mn + n2) x + n3 ... (1)

(mn + m2) y = (mn − n2) x + m3 ... (2)

Let m1 and m2 be the slopes of these lines.

m1=mn+n2m2-mn, m2=mn-n2mn+m2

Let θ be the angle between the lines.

Then,

tan θ=m1-m21+m1m2 =mn+n2m2-mn-mn-n2mn+m21+mn+n2m2-mn×mn-n2mn+m2tan θ=mn+n2mn+m2-mn-n2m2-mnm2-mnmn+m2+mn+n2mn-n2tan θ=m2n2+m3n+mn3+m2n2-m3n+m2n2+m2n2-mn3m3n+m4-m2n2-m3n+m2n2-mn3+mn3-n4

tanθ=4m2n2m4-n4

Hence, the acute angle between the lines is tan-14m2n2m4-n4.

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