Question

If a straight line is at a distance of 10 units from the origin and the perpendicular drawn from the origin to it makes an angle \( \frac{\pi}{4} \) with the negative X-axis in the negative direction, then the equation of that line is

• A. \( x + y + 10\sqrt{2} = 0 \)
• B. \( x - y - 10\sqrt{2} = 0 \)
• C. \( x + y - 10\sqrt{2} = 0 \)
• D. \( x - y + 10\sqrt{2} = 0 \)

Hint:

Use the normal form of a straight line: \( x\cos\theta + y\sin\theta = p \), where \( p \) is the perpendicular distance from the origin and \( \theta \) is the angle the perpendicular makes with the positive X-axis.

Answer 1

The Correct Option is D

Step 1: The perpendicular from the origin to the line makes an angle \( \frac{\pi}{4} \) with the **negative X-axis**, i.e., the angle with the **positive X-axis** is \( \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
So, the normal form of the line is: \[ x \cos\left(\frac{3\pi}{4}\right) + y \sin\left(\frac{3\pi}{4}\right) = p \] Step 2: \( p = 10 \), the perpendicular distance from the origin. \[ x . \left(-\frac{1}{\sqrt{2}}\right) + y . \left(\frac{1}{\sqrt{2}}\right) = 10 \Rightarrow \frac{-x + y}{\sqrt{2}} = 10 \Rightarrow -x + y = 10\sqrt{2} \Rightarrow x - y + 10\sqrt{2} = 0 \]