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I saw the answer to the same doubt I had the other day but can anyone please explain it to me more simply?

Let there be three lines given by the equations: $$ \begin{cases} a_1x+b_1y+c_1=0 \\ a_2x+b_2y+c_2=0\\ a_3x+b_3y+c_3=0\\ \end{cases} $$ Now, a more direct way to prove concurrency of these lines is by the determinant method .

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

Given three lines

  • L1≡a1x+b1y+c1=0
  • L2≡a2x+b2y+c2=0
  • L3≡a3x+b3y+c3=0

are concurrent if. there exist constants λ1,λ2,λ3 not all equal to zero such that **λ1L1+λ2L2+λ3L3=0 or, equivalently: $$ λ_1(a_1x+b_1y+c_1)+λ_2(a_2x+b_2y+c_2)+λ_3(a_3x+b_3y+c_3)=0 $$ I want to ask why is this condition true ? I mean why is it that when the above expression is equal to zero the lines are concurrent?

Matti P.
  • 6,012
Jay Agarwal
  • 11
  • Did you try vector algebra? – Arjun Jun 12 '20 at 11:44
  • i don't know much vector algebra – Jay Agarwal Jun 12 '20 at 12:22
  • The vanishing of the combined eqn requires that the combined coeffs of $x$, the combined coeffs of $y$, and the combined constants each vanish. This gives a linear system in the $\lambda_i$: $$\begin{bmatrix}a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3\end{bmatrix}\begin{bmatrix}\lambda_1\ \lambda_2\ \lambda_3\end{bmatrix}=0$$ This has a non-trivial soln only if the coeff matrix isn't invertible, which requires that its determinant vanishes. But that coeff matrix is the transpose of the lines' coeff matrix, making your condition equivalent to the determinant condition for the line eqns. – Blue Jun 12 '20 at 12:34

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