In the previous post Proof for the equation of a line passing through the intersection of two lines(family of lines). I have looked at derivation which seems to be oversimplified so I am still confused.

Let $C(x_0,y_0)$ be the common point $L_1 \cap L_2$. Thus

$$\begin{cases}a_1x_0 + b_1y_0+ c_1 =0\\a_2x_0 + b_2y_0+ c_2 =0\end{cases}$$

By difference with the initial equations, we obtain the new equivalent equations:

$$\tag{1}\begin{cases}a_1(x-x_0) + b_1(y-y_0)=0 \ \ (L_1)\\a_2(x-x_0) + b_2 (y-y_0)=0 \ \ (L_2)\end{cases} \ \ \implies \ \underbrace{(a_1+Ka_2)(x-x_0) + (b_1+Kb_2)(y-y_0)=0}_{\text{line} (L_1+KL_2)}$$

From equation 1 why do we add a constant $K$ to line $L_2$ and what is its significance?

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    $\begingroup$ Wouldn't this question best be asked in a comment to the answer that describes $K$? Alerting the answerer that the response is unclear will allow the answerer to improve that response. $\endgroup$ – Blue Dec 30 '18 at 12:10
  • $\begingroup$ @Blue It would be a better idea but SE has limit for users to comment(minimum 50 reputations ) and I am short of it. $\endgroup$ – pranjal verma Dec 30 '18 at 13:42
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    $\begingroup$ I added a comment to that answer to direct the answerer here. $\endgroup$ – Blue Dec 30 '18 at 13:57
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    $\begingroup$ We were supposed to prove that any lines passing through the intersection of the two lines has an equation of the form $L_1 + KL_2 = 0.$ (Which is not strictly true; what is true is that every such line except $L_2$ has such an equation.) At some time during the proof we were going to have to refer to that equation; how can you prove something that you never even mention? $\endgroup$ – David K Dec 30 '18 at 19:47

Maybe explaining it with vectors will be clearer.

(see figure below)

A line $L_{P_0,N}$ can be defined by a point $P_0=(x_0,y_0)$ and a normal (= orthogonal) vector $\overrightarrow{N}=(a,b)$ in the following way :

$$M=(x,y) \in L_{P_0,N} \ \ \iff \ \ \overrightarrow{P_0M}=\binom{x-x_0}{y-y_0} \perp \overrightarrow{N}=\binom{a}{b} \ \ \iff \ \ $$


Thus if you have two straight lines passing through the same $(x_0,y_0)$ with resp. normal vector $\binom{a_1}{b_1}$ and $\binom{a_2}{b_2}$, taking now the normal vector $\binom{a_1}{b_1}+K \binom{a_2}{b_2}=\binom{a_1+Ka_2}{b_1+Kb_2}$ you will easily be convinced that you get in this way all possible normal vectors, thus all the pencil of lines passing through $P_0$, with the exception of the 2nd line itself.

This is exactly what means the implication I had given, and that you have reproduced above ; indded, it can be written in the following way:

$$\binom{a_1}{b_1} \perp \binom{x-x_0}{y-y_0} \ \ \ \ \ \text{and} \ \ \ \ \ \binom{a_2}{b_2} \perp \binom{x-x_0}{y-y_0} \implies $$

$$\left(\binom{a_1}{b_1}+K\binom{a_2}{b_2}\right) \perp \binom{x-x_0}{y-y_0} $$

enter image description here

  • $\begingroup$ How do you come with the formulae $\binom{a_1}{b_1} + K\binom{a_2}{b_2} = \binom{a_1 + Ka_2}{b_1 + Kb_2}$,I don't know much about vectors but I assume that $\binom{a}{b}$ is the slope of the orthogonal line. $\endgroup$ – pranjal verma Dec 31 '18 at 9:11
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    $\begingroup$ First sentence : it is a classical linear combination of vectors. Second sentence : $\binom{a}{b}$ is a vector, not a slope. The slope of the corresponding line (with equation $a(x-x_0)+b(y-y_0)=c \iff y=(-b/a)x+...$) is $-b/a$. $\endgroup$ – Jean Marie Dec 31 '18 at 9:18
  • $\begingroup$ Can you show $\binom{a_1}{b_1}+K \binom{a_2}{b_2}=\binom{a_1+Ka_2}{b_1+Kb_2}$ in the form of slope of the orthogonal lines? $\endgroup$ – pranjal verma Dec 31 '18 at 9:21

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