Question

Let \( P(x, y) \) lie on one of the lines: \[ \sqrt{3}x - y + 2 = 0 \quad \text{or} \quad \sqrt{3}x + y - 2 = 0 \] and suppose it is at a distance 5 units from their point of intersection. What is the distance from \( (0, 0) \) to the foot of the perpendicular from \( P \) onto the \( y \)-axis?

• A. \( 2 + \frac{5\sqrt{3}}{2} \)
• B. \( \frac{5\sqrt{3}}{2} \)
• C. \( 2 \)
• D. \( 2 - \frac{5\sqrt{3}}{2} \)

Hint:

To find coordinates satisfying distance conditions, substitute line equations into the distance formula, solve, and compute projection.

Answer 1

The Correct Option is A

Lines: \[ L_1: \sqrt{3}x - y + 2 = 0 \quad L_2: \sqrt{3}x + y - 2 = 0 \] Add both to find intersection point: \[ 2\sqrt{3}x = 0 \Rightarrow x = 0,\ y = 2 \Rightarrow \text{Point of intersection is } (0, 2) \] Point \( P \) is 5 units away from this and lies on either line. Try line \( \sqrt{3}x - y + 2 = 0 \Rightarrow y = \sqrt{3}x + 2 \) Use distance formula between \( P(x, y) \) and \( (0, 2) \): \[ \begin{align} \sqrt{(x)^2 + (y - 2)^2} = 5
\Rightarrow (y - 2)^2 + x^2 = 25
\Rightarrow (\sqrt{3}x)^2 + x^2 = 25 \Rightarrow 3x^2 + x^2 = 25 \Rightarrow x^2 = \frac{25}{4}
\Rightarrow x = \pm\frac{5}{2}
\Rightarrow y = \sqrt{3} \cdot \frac{5}{2} + 2 = 2 + \frac{5\sqrt{3}}{2} \] Now perpendicular from \( P \) to \( y \)-axis is horizontal → distance is \( |x| = \frac{5}{2} \) So distance from origin to foot of perpendicular is: \[ \sqrt{x^2 + 0^2} = \frac{5}{2},\quad \text{but foot is at } x = 0,\ y = y \Rightarrow \text{Distance from (0,0) to } (0,y) = |y| = 2 + \frac{5\sqrt{3}}{2} \]