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Let there be three lines given by the equations: $$a_1x+b_1y+c_1=0\\ a_2x+b_2y+c_2=0\\ a_3x+b_3y+c_3=0$$'

Now, a more direct way to prove concurrency of these lines is just proving $$\begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}=0$$

But, I came across another condition for proving concurrency which I am not able to understand. It goes as follows:

Given three lines $$L_1\equiv a_1x+b_1y+c_1=0\\ L_2\equiv a_2x+b_2y+c_2=0\\ L_3\equiv a_3x+b_3y+c_3=0$$ are concurrent iff. there exist constants $\lambda_1,\lambda_2,\lambda_3$ not all equal to zero such that $$\lambda_1L_1+\lambda_2L_2+\lambda_3L_3=0$$ or, equivalently: $$\lambda_1(a_1x+b_1y+c_1)+\lambda_2(a_2x+b_2y+c_2)+\lambda_3(a_3x+b_3y+c_3)=0$$

What does the last equation mean? What are $x$ and $y$ in it? If $(x,y)$ represent the point of concurrency of the lines, won't every real number $\lambda$ satisfy the above equality? Can someone explain this to me?

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For simplicity let's call the new criterion you've found $\mathcal C$.

To say that the three lines are concurrent iff $\mathcal C$ means that starting from the definition of a concurrent system of lines we should be able to derive $\mathcal C$ and starting from $\mathcal C$ we should be able to derive the fact that the lines are concurrent.

Proof that lines are concurrent $\implies \mathcal C$:

Let $$a_1x+b_1y+c_1=0\\ a_2x+b_2y+c_2=0\\ a_3x+b_3y+c_3=0$$ such that $\begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}=0$. We know that

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} = 0 \\ \iff \lambda_1(a_1,b_1,c_1)+\lambda_2(a_2, b_2, c_2) + \lambda_3(a_3, b_3, c_3)=0 \text{ has a nontrivial solution}^\dagger$$

That is, iff you can find some numbers $\lambda_1, \lambda_2,\lambda_3$, not all zero, such that each of the following are simultaneously true:

$$\lambda_1a_1+\lambda_2a_2 + \lambda_3a_3 = 0 \\ \lambda_1b_1+\lambda_2b_2 + \lambda_3b_3 = 0 \\ \lambda_1c_1+\lambda_2c_2 + \lambda_3c_3 = 0$$

Now multiply both sides of that first equation by $x$, the second by $y$, and then add them all together to get:

$$(\lambda_1a_1+\lambda_2a_2 + \lambda_3a_3)x + (\lambda_1b_1+\lambda_2b_2 + \lambda_3b_3)y + \lambda_1c_1+\lambda_2c_2 + \lambda_3c_3 = 0$$

Rearranging and factoring out the $\lambda$'s, we get

$$\lambda_1(a_1x+b_1y+c_1) + \lambda_2(a_2x+b_2y+c_2) + \lambda_3(a_3x+b_3y+c_3) = 0 \\ \lambda_1L_1 + \lambda_2L_2 + \lambda_3L_3 = 0$$

Hence the lines are concurrent $\implies \mathcal C$.

I'm not going to do the other implication for you. See if you can do it yourself.


$\dagger$: Hopefully you've seen this before, but if not you're going to have to learn it. If you're at a university, see if you can download chapter 2 of this book for more details (specifically look at theorem 2.4). If you don't have access, see if you can glean out the important facts from the matrix invertibility theorem.

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  • $\begingroup$ Okay, I see what that means. Maybe if you could tell why and how that works... $\endgroup$ – codetalker Oct 22 '16 at 13:33

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