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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
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