Find the value of the acute angle between the lines 3x-y+1=0 and x-2y+1=0.
Given lines are-
3x−y+1=0⇒y=3x+13x-y+1=0\Rightarrow y=3x+13x−y+1=0⇒y=3x+1
x−2y+1=0⇒y=x2+12x-2y+1=0\Rightarrow y=\dfrac{x}{2}+\dfrac{1}{2}x−2y+1=0⇒y=2x+21
Slope of the lines are-
m1=3,m2=12m_1=3,m_2=\dfrac{1}{2}m1=3,m2=21
So Angle between two lines-
tanθ=m1−m21+m2m1tan\theta=\dfrac{m_1-m_2}{1+m_2m_1}tanθ=1+m2m1m1−m2
=3−0.51+3(0.5)=2.52.5=1=\dfrac{3-0.5}{1+3(0.5)}\\[9pt]=\dfrac{2.5}{2.5}=1=1+3(0.5)3−0.5=2.52.5=1
θ=45∘\theta=45^{\circ}θ=45∘
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