Question

The angle made by a line \(L\) with positive X-axis measured in the positive direction is \(\frac{\pi}{6}\) and the intercept made by \(L\) on Y-axis is negative. If \(L\) is at a distance 5 units from the origin, then the perpendicular distance from the point \(\left(1,-\sqrt{3}\right)\) to the line \(L\) is?

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

Use the point-to-line distance formula carefully with slope-intercept form and consider sign of intercept.

Answer 1

The Correct Option is D

Slope \(m = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\). Line equation: \(y = mx + c\) with \(c<0\). Distance from origin to line is \[ \frac{|c|}{\sqrt{1 + m^2}} = 5 \implies |c| = 5 \times \sqrt{1 + \frac{1}{3}} = 5 \times \frac{2}{\sqrt{3}} = \frac{10}{\sqrt{3}}. \] Since \(c<0\), \(c = -\frac{10}{\sqrt{3}}\). Distance from point \((1, -\sqrt{3})\) to line: \[ d = \frac{|y_1 - m x_1 - c|}{\sqrt{1 + m^2}} = \frac{\left| -\sqrt{3} - \frac{1}{\sqrt{3}} . 1 + \frac{10}{\sqrt{3}} \right|}{\frac{2}{\sqrt{3}}} = 3. \]