Question

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

Answer 1

Given lines are-

$3x-y+1=0\Rightarrow y=3x+1$

$x-2y+1=0\Rightarrow y=\dfrac{x}{2}+\dfrac{1}{2}$

Slope of the lines are-

$m_1=3,m_2=\dfrac{1}{2}$

So Angle between two lines-

$tan\theta=\dfrac{m_1-m_2}{1+m_2m_1}$

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

$\theta=45^{\circ}$