Question

Suppose \(P(x,y)\) lying on \[ \sqrt{3}x-y+2=0 \] or \[ \sqrt{3}x+y-2=0 \] is at a distance of \(5\) units from their point of intersection. Then the distance from \((0,0)\) to the foot of the perpendicular of \(P\) onto the \(y\)-axis is

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

Hint:

If a point moves a distance \(r\) along a line making angle \(\theta\) with the positive \(x\)-axis, then its vertical change is \[ r\sin\theta. \]

Answer 1

The Correct Option is A

Step 1: Find the point of intersection of the two lines.
The given lines are \[ \sqrt{3}x-y+2=0 \] and \[ \sqrt{3}x+y-2=0 \] From the first line, \[ y=\sqrt{3}x+2 \] From the second line, \[ y=2-\sqrt{3}x \] Equating, \[ \sqrt{3}x+2=2-\sqrt{3}x \] \[ 2\sqrt{3}x=0 \] \[ x=0 \] Then, \[ y=2 \] So, the point of intersection is \[ (0,2) \]

Step 2: Use the distance condition.
Point \(P\) lies on either line and is at distance \(5\) from \((0,2)\).
The lines make an angle of \(60^\circ\) with the positive \(x\)-axis.
Therefore, the vertical change in moving \(5\) units along the line is \[ 5\sin 60^\circ \] \[ =5\cdot \frac{\sqrt{3}}{2} \] \[ =\frac{5\sqrt{3}}{2} \]

Step 3: Find the required distance on the \(y\)-axis.
Since the point of intersection has \(y\)-coordinate \(2\), the required distance from \((0,0)\) to the foot of perpendicular on the \(y\)-axis is \[ 2+\frac{5\sqrt{3}}{2} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2+\frac{5\sqrt{3}}{2}} \]