Question:
The perpendicular drawn from the origin to the straight line $\sqrt{3}x + y - 24 = 0$ makes an angle $\alpha$ with the positive direction of x-axis. Then $\alpha$ is equal to:
The perpendicular drawn from the origin to the straight line $\sqrt{3}x + y - 24 = 0$ makes an angle $\alpha$ with the positive direction of x-axis. Then $\alpha$ is equal to:
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Perpendicular slope = negative reciprocal of original slope.
Updated On: Apr 24, 2026
- $120^\circ$
- $45^\circ$
- $135^\circ$
- $60^\circ$
- $30^\circ$
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The Correct Option is
Solution and Explanation
Concept:
• Slope of line: $m = -\frac{a}{b}$
• Slope of perpendicular = negative reciprocal
Step 1: Find slope of given line
\[ \sqrt{3}x + y - 24 = 0 \Rightarrow y = -\sqrt{3}x + 24 \] \[ m_1 = -\sqrt{3} \]
Step 2: Find slope of perpendicular
\[ m_2 = \frac{1}{\sqrt{3}} \]
Step 3: Find angle
\[ \tan\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = 30^\circ \] Final Conclusion:
\[ \alpha = 30^\circ \]
• Slope of line: $m = -\frac{a}{b}$
• Slope of perpendicular = negative reciprocal
Step 1: Find slope of given line
\[ \sqrt{3}x + y - 24 = 0 \Rightarrow y = -\sqrt{3}x + 24 \] \[ m_1 = -\sqrt{3} \]
Step 2: Find slope of perpendicular
\[ m_2 = \frac{1}{\sqrt{3}} \]
Step 3: Find angle
\[ \tan\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = 30^\circ \] Final Conclusion:
\[ \alpha = 30^\circ \]
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