if the algebriac sum of distances of points(2,1);(3,2) and (-4,7) from the line y =mx+c is zero;then this line will pass through a fixed point .find the co-ordinate of the fixed point

let that line be  ax+ by + c= 0 

then acc to question  ( 2a + b + c ) + ( 3a+2b+c)+(-4a+7b+c) /√a²+b²+c²  = 0  by putting the values in the eq 

on solving it we get  a +10b +3c=0  or, c= -a/3 - 10b/3

so the eq becomes  ax +by -a/3-10b/3 = 0   or   a(3x - 1 ) + b (3y-10) = 0 

so the intersection point of the lines   3x-1=0  and   3y-10=0  is the given point