Question:
A straight line L through the point \((3,-2)\) is inclined at an angle of \(60^\circ\) to the line \(\sqrt{3}\,x+y=1\). If L also intersects the \(x\)-axis, then the equation of L is :
A straight line L through the point \((3,-2)\) is inclined at an angle of \(60^\circ\) to the line \(\sqrt{3}\,x+y=1\). If L also intersects the \(x\)-axis, then the equation of L is :
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Find the slopes making 60 degrees with the given line; reject the horizontal one because L must meet the x-axis.
Updated On: Nov 11, 2026
- \(y+\sqrt{3}\,x+2-3\sqrt{3}=0\)
- \(y-\sqrt{3}\,x+2+3\sqrt{3}=0\)
- \(\sqrt{3}\,y-x+3+2\sqrt{3}=0\)
- \(\sqrt{3}\,y+x-3+2\sqrt{3}=0\)
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The Correct Option is B
Solution and Explanation
Step 1: The given line has slope \(-\sqrt{3}\). Let L have slope m: \(\tan 60^\circ=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|=\sqrt{3}\).
Step 2: \(m+\sqrt{3}=\sqrt{3}(1-\sqrt{3}m)\Rightarrow m=0\); or \(m+\sqrt{3}=-\sqrt{3}(1-\sqrt{3}m)\Rightarrow m=\sqrt{3}\).
Step 3: \(m=0\) gives a line parallel to the x-axis (never meets it), so \(m=\sqrt{3}\).
Step 4: \(y+2=\sqrt{3}(x-3)\Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0\).
Answer: (2) \(y-\sqrt{3}\,x+2+3\sqrt{3}=0\)
Step 2: \(m+\sqrt{3}=\sqrt{3}(1-\sqrt{3}m)\Rightarrow m=0\); or \(m+\sqrt{3}=-\sqrt{3}(1-\sqrt{3}m)\Rightarrow m=\sqrt{3}\).
Step 3: \(m=0\) gives a line parallel to the x-axis (never meets it), so \(m=\sqrt{3}\).
Step 4: \(y+2=\sqrt{3}(x-3)\Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0\).
Answer: (2) \(y-\sqrt{3}\,x+2+3\sqrt{3}=0\)
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