3x + y + 12 = 0 and x + 2y – 1 = 0

Given:

The equations of the lines are

3x + y + 12 = 0 … (1)

x + 2y − 1 = 0 … (2)

Let m_{1} and m_{2} be the slopes of these lines.

m_{1} = -3, m_{2} = \( \dfrac{-1}{2}\)

Let θ be the angle between the lines.

Then, by using the formula

= \(\dfrac{ (-3 + \dfrac{1}{2}) }{ (1 + \dfrac{3}{2})}\)

= 1

So,

θ =\( \dfrac{\pi}{4}\)or 45°

The acute angle between the lines is 45°

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