Question:medium

Find angles between the lines \(\sqrt3x+y=1\) and \(x+\sqrt3y=1.\)

Updated On: Jan 22, 2026
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Solution and Explanation

The given lines are \(\sqrt3x+y=1\) and \(x+\sqrt3y=1.\)

\(y = -\sqrt3x + 1 … (1)\) and \(y = \frac{-1}{\sqrt3x} +\frac{ 1}{\sqrt3} …. (2)\)

The slope of line (1) is  \(m_1 = -\sqrt3\) , while the slope of line (2) is  \(m_2 =\frac{ -1}{\sqrt3}\)
The acute angle i.e., \(θ\) between the two lines is given by

\(Tanθ=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\)

\(Tanθ=\left|\frac{-\sqrt3+\frac{1}{\sqrt3}}{1+(-\sqrt3)(\frac{-1}{\sqrt3})}\right|\)

\(Tanθ=\left|\frac{\frac{-3+1}{\sqrt3}}{1+1}\right|\)

\(=\left|\frac{-2}{2\times\sqrt3}\right|\)

\(Tanθ=\frac{1}{\sqrt3}\)

\(θ=30º\)
Thus, the angle between the given lines is either \( 30° \)or \(180° - 30° = 150°\).

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