#1 Assume $\mathbf{a_1x+b_1y+c_1=0}$ and $\mathbf{a_2x+b_2y+c_2=0}$ two linear equations. If we add these two equations we get $$\mathbf{(a_1+a_2)x+(b_1+b_2)y+(c_1+c_2)=0}$$ $$\mathbf{y=\frac{-(a_1+a_2)}{b_1+b_2}x+\frac{-(c_1+c_2)}{b_1+b_2}}$$ $$\mathbf{m=\frac {-(a_1+a_2)}{b_1+b_2}}$$ #2 Now we can write the first and second equation respectively as $\mathbf{y=\frac{-a_1}{b_1}x-\frac{c_1}{b_1}}$ and $\mathbf{y=\frac{-a_2}{b_2}x-\frac{c_2}{b_2}}$. Add this two equations like before and we get, $$\mathbf{2y=\left(\frac{-a_1}{b_1}+\frac{-a_2}{b_2}\right)x-\left(\frac{c_1}{b_1}+\frac{c_2}{b_2}\right)}$$ $$\mathbf{y=-\frac{(a_1b_2+a_2b_1)}{2b_1b_2}x-\frac{1}{2}\left(\frac{c_1}{b_1}+\frac{c_2}{b_2}\right)}$$ So, $$\mathbf{m=-\frac{(a_1b_2+a_2b_1)}{2b_1b_2}}$$

I just added two same equations here in two different ways. So I should get the same equation for both cases. But apparently the slopes are different here. I believe Somewhere I have made a very silly mistake. I took my time and went through it carefully but unfortunately couldn't find the mistake. Can you please show me where I made the mistake?

  • $\begingroup$ You have a different representation of the same line. Note that the constant term in both equations is also different, not just the slope. $\endgroup$
    – fosho
    Jan 7, 2016 at 9:52
  • $\begingroup$ Yes. that was my question. Just because my representation is different the result shouldn't be different, don't you think? All I did was adding two equations. No matter how I add two equations the output must be same I think $\endgroup$ Jan 7, 2016 at 9:58
  • $\begingroup$ So do you think $1-\frac{1}{x}$ and $\frac{x-1}{x}$ are different? $\endgroup$
    – fosho
    Jan 7, 2016 at 10:01
  • $\begingroup$ Definitely not. But i don't get how your example is related to my question. Both value m was supposed to be the same. But it's not $\endgroup$ Jan 7, 2016 at 10:08
  • $\begingroup$ You're observing this difference because you've divided through, in the second part of your question, by two potentially different values: $b_1$ and $b_2$. $\endgroup$
    – amcerbu
    Jan 7, 2016 at 13:00

3 Answers 3


The problem is, that in #2 you divide each of the equations by $b$ to get them in slope-intercept form. So in #1 you get the equation $eq_1+eq_2$, whereas in #2 you get $\frac{eq_1}{b_1}+\frac{eq_2}{b_2}$. Clearly, those shouldn't give the same equations.


Hint take for eg equations $2x+3y+5=0,3x+4y+7=0$ add it in ways you have done you wont get similar lines hence the slopes and constants are different . This is because of the different representation of same line .

  • 1
    $\begingroup$ I'm not the OP, but I'm confused about this matter too. What do you mean by "different representation of the same line"? If coordinate axes are fixed then doesn't a line correspond to a unique representation? $\endgroup$
    – user160738
    Jan 7, 2016 at 12:28
  • $\begingroup$ The representation $y=mx+c$ is slope intercept form so there are infinite lines goin throught same point but with different intercepts on axes. While when we add two equTions its useless as any point satisfying thise two will surely satisfy third equation.but two lines if not the same line tgen they will have same point of intersection and will have same intercepts hence the difference $\endgroup$ Jan 7, 2016 at 12:38

Summary: when we 'add' two lines as linear equations to get a third, there are 3 cases:

  • If they're parallel and equal, the result will be the same line
  • If they're parallel and distinct, the result will be parallel, and we can get any parallel line we like except for the first two by multiplying the original equations by non-zero scalars.
  • If they intersect, the result will pass through the point of intersection and we can give it any slope we like except for the slopes of the first two like by multiplying the original by non-zero scalars.
  • When you're doing your two sets of calculations, you're effectively scaling one pair of equations to change their $y$ coefficients to 1 (so you get $y=\cdots$). This will give you a different result slope unless the original lines were parallel or you applied the same scaling factor to both (i.e. the original equations had the same $y$ coefficients to start with).

TLDR explanation: When we add any two equations $a=b$ and $c=d$ to get $a+c=b+d$, we do so on the basis of consistency. It's a little deduction: if $a=b$ and $c=d$ then $a+c=b+d$. When adding two linear equations to get a third, it means solutions to the first two (values of variables which make both the first two equations true at the same time) are preserved in the third (the same values make the third equation true). It also means the third lines intersects either of the first two only at solution points to the first two. That's why parallel lines give parallel lines (distinct means no solutions means third line doesn't intersect, i.e. is parallel; identical means infinite solutions means third line the same) and lines that intersect at a point give another (different) line that intersects at that point.

Example: parallel lines $y=x$ and $y=x+1$. We add equations to get $2y=2x+1$ or $y=x+\frac{1}{2}$, which is parallel as expected. If we multiply the equations by non-zero scalars we get $k_1y=k_1x$ and $k_2y=k_2x+k_2$ which sum to $(k_1+k_2)y=(k_1+k_2)x+k2$, or $y=x+\dfrac{k_2}{k_1+k_2}$ and we can make the $y$-intercept any number $c$ we like except $0$ or $1$ by picking $k_2=c, k_1=1-k_2$.

Example: lines $y=x$ and $y=-x+2$ intersect at $(1,1)$. The equations sum to $2y=2$ or $y=1$ which passes through the solution point $(1,1)$ as expected and has slope $0$. Adding non-zero scalar multiples of the original equations gives $(k_1+k_2)y=(k_1-k_2)x+2k_2$, which has slope $\dfrac{k_1-k_2}{k_1+k_2}$ which we can set to any number $m$ we like except $1$ or $-1$ by picking $k_1=\dfrac{1+m}{2}, k_2=\dfrac{1-m}{2}$, or undefined (vertical lines) by picking $k_1=1,k_2=-1$.


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