Arun
Last Activity: 6 Years ago
Let the line passing through (a,b) be,
y= mx-am+b, ….....................................(i)
Now a line perpendicular to this passing through origin is,
y=-x/m,.......................................................(ii)
let the feet of the perpendicular be (h,k)
From (i)
k= mh-am+b,.....................................(iii)
From (ii)
m= -h/k,
Replacing this in (iii)
k= (-h/k)*h + a(h/k) +b,
hence, k2=-h2+ah+bk,
So h2+k2= ah+bk,
Hence the locus is x2+y2=ax+by
But in question, a = 2, b = 3
Hence
Locus
x² + y² - 2x -3y = 0
This will be a circle
dia = sqrt (1² + (3/2)²)
= sqrt (13/4)
Regards
Arun (askIITians forum expert)
