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 m1 and m2 be the slopes of these lines.
m1 = -3, m2 = \( \dfrac{-1}{2}\)
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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