Question

The lines \[ L_1:2x+y+1=0 \] and \[ L_2:x-2y+4=0 \] intersect at \(A\). Let \(P\) be a point at a distance \(5\) units from \(L_1=0\) and \(\alpha\) units from \(L_2=0\). If \(M\) and \(N\) are the feet of the perpendiculars from \(P\) on the lines \(L_1=0\) and \(L_2=0\) respectively, and the area of the quadrilateral \(AMPN\) is \(25\) sq. units, then the point \(P\) lies on the line

• A. \[ x+3y-3=0 \]
• B. \[ 3x-y+12=0 \]
• C. \[ x+3y=0 \]
• D. \[ 3x-y=0 \]

Hint:

If a point is equidistant from two intersecting lines, it lies on one of their angle bisectors. Use \[ \frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}} \] to obtain the angle bisectors directly.

Answer 1

The Correct Option is A