The locus of the mid - points of the perpendiculars drawn from points on the line, x=2y to the line x=y is:
(A)7x−5y=0
(B)3x−2y=0
(C)2x−3y=0
(D)5x−7y=0✓ Correct
Answer: (D)
Solution
Take a point (2t,t) on the line x=2y and drop a perpendicular onto x=y, meeting it at (s,s). Since that perpendicular has slope −1, s−2ts−t=−1, which gives s=23t. The midpoint (h,k) of the segment is h=22t+3t/2=47t and k=2t+3t/2=45t. Eliminating t through 7h=5k gives the locus 5x−7y=0.
JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option