Why does $Ax+By+C+\lambda(A'x+B'y+C')=0$ represent the set of lines that pass through the intersection of $Ax+By+C=0$ and $A'x+B'y+C'=0$

Question

For example, given $\color{green}{l_1:5x-2y-8=0}$ and $\color{blue}{l_2:3x+8y-8=0}$,

We can compute the set of lines that pass through the intersection of $l_1$ and $l_2$

$$5x-2y-8+\lambda(3x+8y-8)=0$$ $$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$ for any $\lambda\in\Bbb{R}$

$$\lambda$$	$$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$
$$-1$$	$$x-5y=0$$
$$1$$	$$4x+3y-8=0$$
$$2$$	$$11x+14y-24=0$$

Why does this method work?

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The closest thread I could find is, A general circle through the intersection points of line $l$ and circle $S_1$ has the form $S_1+\lambda L$. What is the significance of $\lambda$?.

For example if we want to find lines through the point of intersection of 3x+4y+5=0 and 2x+y+4=0 . The required lines would be obtained by substituting different values of λ in 3x+4y+5+λ(2x+y+4)=0

The accepted answer is,

Let us take up the case of lines first. Let $L_1(x,y)$ and $L_2(x,y)$ be two lines which intersect at $(a,b)\\$. Thus $$L_1(a,b)=0\\L_2(a,b)=0$$Now let $L_3(x,y)$ be another line such that $$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$$Now, if we are able to show that $L_3$ passes through $(a,b)$,i.e. the intersection point of $L_1$ and $L_2$, our job will be complete.To do this we put $(a,b)$ in our expression for $L_3$ $$L_3(a,b)=L_1(a,b)+\lambda L_2(a,b)$$ $$\Rightarrow L_3(x,y)=0+\lambda .0$$ $$\Rightarrow L_3(x,y)=0$$So as you can see, for any value of $\lambda$, our line $L_3$ always passes through the intersection of lines $L_1$ and $L_2\\$. You can the same with any two curves(e.g. two circles) .

What I understood from this answer is,

• Consider the lines $L_1(x, y)$ and $L_2(x, y)$ which intersept at $(a, b)$ such that $L_1(a, b)=0$ and $L_2(a, b)=0$
• Assume that $L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$
• Then this means that $L_3$ passes through $(a, b)$

I do not understand how this proves that $L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$ spans a set of distinct lines that pass through $(a, b)$.

In the case of circles of the form $x^2+y^2+Dx+Ey+F=0$ and $x^2+y^2+D'x+E'y+F'=0$ that intersect and are not concentric, we can't have $\lambda=-1$ because we would get their radical axis, not another circle.

I'm looking for a proof by deduction if possible (as opposed to assumption).

Answer 1

One explanation is to show that for any point $(p,q)$ outside the line $L_2(x,y)=0$, we can set $\lambda$ so that the the point is on the line $L_1(x,y)+\lambda L_2(x,y)=0$.

This is possible because by the assumption $L_2(p,q)\neq 0$ so we can set $\lambda=-L_1(p,q)/L_2(p,q).$

Edit: I mean that we can obtain the line $L_1(x,y)+\lambda L_2(x,y)=0$ passing through $(p,q)$ by setting $\lambda=-L_1(p,q)/L_2(p,q).$

Answer 2

Let's say that $$L_1 : \{Ax+By+C=0\} \Longrightarrow \vec{L_1} =(A, B, C)$$ $$L_2 : \{A'x+B'y+C'=0\} \Longrightarrow \vec{L_2} =(A', B', C')$$ The intersection point can be found by solving the system by Cramer rule: $$\begin{pmatrix}L_1 \\ L_2 \end{pmatrix} = \begin{pmatrix}A&B&C \\ A'&B'&C'\end{pmatrix}$$

$$x=\frac{CB'-BC'}{AB'-BA'} \qquad y=\frac{AC'-CA'}{AB'-BA'}$$

Doing $R_2 = L_1 + \lambda(L_2) = L_3$ $$\begin{pmatrix}L_1 \\ L_3 \end{pmatrix} = \begin{pmatrix}A&B&C \\ A+\lambda A'&B+\lambda B'&C+\lambda C'\end{pmatrix}$$

$$x=\frac{C(B+\lambda B')-B(C+\lambda C')}{A(B+\lambda B')-B(A+\lambda A')} \qquad y=\frac{A(C+\lambda C')-C(A+\lambda A')}{A(B+\lambda B')-B(A+\lambda A')}$$

$$x=\frac{\lambda CB'-\lambda BC'}{\lambda AB'-\lambda BA'} \qquad y=\frac{\lambda AC'-\lambda CA'}{\lambda AB'-\lambda BA'}$$ $$x=\frac{CB'-BC'}{AB'-BA'} \qquad y=\frac{AC'-CA'}{AB'-BA'}$$

Which means that $L_1 \cap L_2 = L_1 \cap L_3 $