Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0

Asked by Sakshi | 1 year ago |  164

##### Solution :-

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°

Answered by Aaryan | 1 year ago

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