If each of the points (x1,4),(−2,y1) lies on the line joining the points (2, –1), (5, –3), then the point P(x1,y1) lies on the line

Equation of line passing through (2,–1) and (5, –3) is y+1=(−3+15−2)(x−2)                         ⇒3y+3=−2x+4                         ⇒        2x+3y=1 ….. …(i)(x1,4)  and  (−2,y1) lie  on  Eq.  (i) Then,   2x1+12=1⇒  x1=−112 and  −4+3y1=1⇒   y1=53 Hence,   point  P(−112,53)  lies  an  2x+6y+1=0 .