Question #194314

Find the value of the acute angle between the lines 3x-y+1=0 and x-2y+1=0.


1
Expert's answer
2021-05-25T08:01:01-0400

Given lines are-

3xy+1=0y=3x+13x-y+1=0\Rightarrow y=3x+1


x2y+1=0y=x2+12x-2y+1=0\Rightarrow y=\dfrac{x}{2}+\dfrac{1}{2}


Slope of the lines are-


m1=3,m2=12m_1=3,m_2=\dfrac{1}{2}



So Angle between two lines-


tanθ=m1m21+m2m1tan\theta=\dfrac{m_1-m_2}{1+m_2m_1}


=30.51+3(0.5)=2.52.5=1=\dfrac{3-0.5}{1+3(0.5)}\\[9pt]=\dfrac{2.5}{2.5}=1


θ=45\theta=45^{\circ}


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