The locus of midpoints of the perpendicular drawn from the points on the line  $\mathrm{x}=3\mathrm{y}$ to the line x=y is

detailed solution

Correct option is D

The general point on the line x=3y  is   3h,hThe foot of the perpendicular of 3h,h on the line  x=y is  l,m l−3h1=m−h−1=−3h−h12+12 Simplify l−3h1=−h⇒l=2h  And  m−h−1=−h⇒m=2h Hence the foot of the perpendicular of 3h,h on the line x−y=0 is  2h,2hSuppose that Px1,y1 be any point on the locus.The point  Px1,y1 is midpoint of 3h,h  and  2h,2hIt implies that  x1,y1=5h2,3h2Eliminate h  to get the locus of the point  Px1,y1 Therefore, the locus is 2x5=2y3⇒3x−5y=0