
Shortest distance of a point (x1,y1) from line ax+by=c is d=a2+b2ax1+by1−c Now shortest distance of P(1,2) from 3x +4y=9 is PC=d=32+423(1)+4(2)−9=52 Given that △APB is an equilateral triangle Let ' a ' be its side then PB=a,CB=2a Now, In ΔPCB,(PB)2=(PC)2+(CB)2 (By Pythagoras theorem) a2=(52)2+4a2 a2−4a4=254⇒43a2=254 a2=7516⇒a=7516=534×33=1543