Question

# Find the angle 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+n2 and (mn+m2)y=(mn+n2)x+m3.

Solution

## (i) The given equations are 3x+y+12=0 x+2y−1=0 Writing the equation in the form y=mx+c 3x+y+12=0 y=−3x−12 ⇒m1=−3 Also, x+2y−1=0 2y=1−x y=12−x2 ⇒m2=−12 Angle between the lines tan θ=∣∣m1−m21+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 3x−y+5=0 ⇒y=3x+5 ⇒m1=3 Also, x−3y+1=0 3y=x+1 y=x3+13 ⇒m2=13 Thus angle between the lines is tan θ=∣∣m1−m21+m1m2∣∣ =∣∣∣3−131+313∣∣∣=∣∣∣9−131+1∣∣∣ =∣∣∣832∣∣∣=∣∣86∣∣=43 ⇒θ=tan−1(43) (iii) 3x+4y−7=0 and 4x−3y+5=0 To find angle between the lines, convert the equations in the form y=mx+c 3x+4y−7=0 ⇒4y=−3x+7 y=−34x+74 m1=−34 Also, 4x−3y+5=0 ⇒3y=4x+5 ⇒y=43x+53 ⇒m1=43 The angle between the line is given by tan θ=∣∣m1−m21+m1m2∣∣ =∣∣ ∣∣−34−431+(−3)4(43)∣∣ ∣∣=∣∣∣−34−431−1∣∣∣ ⇒θ=π2 or 90∘ (iv) x−4y=3 and 6x−y=11 To find angle convert the equation in the form y=mx+c x−4y=3 ⇒4y=x−3 y=x4−34 ⇒m1=14 Also, 6x−y=11 y=6x−11 ⇒m2=6 Thus, angle between the lines is tan θ=∣∣m1−m21+m1m2∣∣ =∣∣∣14−61+14×6∣∣∣ =∣∣∣−2341+32∣∣∣=∣∣∣−23452∣∣∣ ⇒θ=tan−1 (2310) (v) (m2−mn)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)m2−mnx+n3(m2−mn) ⇒m1=mn+n2m2−mn Also, y=(mn−n2)nm+m2x+m3nm+m2 ⇒m2=mn−n2nm+m2 Thus, angle between 2 lines is tan θ ⇒tan θ=∣∣m1−m21+m1m2∣∣ =∣∣ ∣∣(mn+n2m2−mn)−(mn−n2nm+m2)1+(mn+n2m2−mn)(mn−n2nm+m2)∣∣ ∣∣ =∣∣m2n2+m3n+n3m+n2m2−m3n+m2n2+n2m2−mn3m3n+m4−m2n2−m3n+m2n2−nm3+mn3−n4∣∣ =∣∣4m2n2m4−n4∣∣ ⇒θ=tan−1∣∣4m2n2m4−n4∣∣

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