# ​If  a+2b+3c = 0  then  a/3+2b/3+c = 0 and comparing with  line ax+ by +v, we get  x = 1/3 & y = 2/3 so there will be a point (1/3, 2/3) from where  each  of the  lines of the form ax + by +c = 0 will pass for the  given  relation between  a. b. c. we can say  if there  exists a linear relation  between  a.b.c then the family of straight  lines of the form of ax + by +c pass  through  a fixed point. If a,b,c are consecutive odd integers then the line ax + by +c= 0 will pass through ?

Dear student

Let a, b and c be three consecutive odd integers as 2k+1,2k+3 and 2k+5

ax+by+c = 0

(2k+1)x+(2k+3)y+(2k+5) = 0

2k(x+y+1)+(x+3y+5) = 0

This equation represents a family of lines passing through the intersection of x+y+1=0 and x+3y+5=0 , irrespective of the value of k.

x+y+1 = 0   ...(1)

x+3y+5 = 0    ...(2)

Subtracting (1) from (2), we get

2y + 4 = 0

y = -2

Now, from (1), we get

x-2+1 = 0

x = 1

Hence fixed point is (1,-2)

Regards

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