The angle between the lines 3x+y=1 and x+3y=1 is

3x+y=1 m1=−3; m2=−13 tanθ=|m1−m21+m1m2| =|−3+131+(−3)(−13)| =|−3+131+1|=|−22×3| =13 θ=30°

The angle between the lines 3x+y=1 and x+3y=1 is

3x+y=1 m1=−3; m2=−13 tanθ=|m1−m21+m1m2| =|−3+131+(−3)(−13)| =|−3+131+1|=|−22×3| =13 θ=30°

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The angle between the lines $\sqrt{3} x + y = 1$ and $x + \sqrt{3} y = 1$ is

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45 °

60 °

30 °

90 °

answer is C.

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Detailed Solution

\sqrt{3} x + y = 1

$m_{1} = - \sqrt{3}; m_{2} = \frac{- 1}{\sqrt{3}}$

$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

$= | \frac{- \sqrt{3} + \frac{1}{\sqrt{3}}}{1 + (- \sqrt{3}) (\frac{- 1}{\sqrt{3}})} |$

$= | \frac{- 3 + 1}{\frac{\sqrt{3}}{1 + 1}} | = | \frac{- 2}{2 \times \sqrt{3}} |$

$= \frac{1}{\sqrt{3}}$

$θ = 30 °$

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The angle between the lines $\sqrt{3} x + y = 1$ and $x + \sqrt{3} y = 1$ is