#
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 ?

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

**
**