comparing vs equating coefficients

Question

Consider two linear equations

 Ax + By + C = 0

 ax + by + c = 0

Now if i proceed in the following way :

  Ax + By + C = ax + by + c    ( Since both are equal to zero )
 (A-a)x + (B-b)y + (C-c) = 0
 (A-a)x + (B-b)y + (C-c) = 0x + 0y + 0

which will lead me to A = a ; B = b ; C = C; assuming non-zero vlues for x and y which is clearly wrong.

My question is why is the above wrong , and could you please clarify what does one mean by comparing coefficients and equating them ?

Answer 1

The big problem here, as I see it, is that you aren't entirely clear on what the equations say. Do they say that

• $A, B, C, a, b, c$ are such that the equations are true for any $x, y, z$?
• $A, B, C, a, b, c$ are fixed (although possibly unknown) and we limit ourselves to values of $x, y, z$ such that both equations are satisfied simultaneously?
• $A, B, C, a, b, c$ are fixed (although possibly unknown) and we study the set of $x, y, z$ which satisfy the first equation and the set of $x, y, z$ which satisfy the second equation and see in what way we may compare the two?
• something else?

How to correctly treat the two equations will depend greatly on what they mean, and until that is clear you can't really put them togeher in any way.

Answer 2

There is not contradiction in your result indeed the following

$$ (A-a)x + (B-b)y + (C-c) = 0$$

is always true for every x and y if and only if the two lines coincides that is $A=a$, $B=b$,$C=c$.

Otherwise you can obtain x or y by one equation, sustitute in the other and find the intersection point.

Answer 3

The equation $$(3-2)x+(4-3)y+(6-5)=0 \\ \implies x + y + 1 = 0$$This equations holds true for all values of $x$ and $y$ at a certain line on 2D plane. So it the coefficients can be something other than $0$ and the equation will still hold.

If you say that equation $$ax+by+c=0$$ holds for all values of $x$ and $y$, then the coefficients have to be zero. But if that isn't the case you can have non-zero coefficients.