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My textbook states:

Any line through the point of intersection of the lines $a_1x +b_1y+c_1=0$ and $a_2x+b_2y+c_2=0 $ can be represented by the equation:

$a_1x +b_1y+c_1+ \lambda(a_2x+b_2y+c_2)=0$ //where $\lambda$ is a parameter.

Now, this theroem (including the $\lambda$ ) is difficult to understand. I am unable to grasp it's concept. I searched for video lectures on Family of Lines but there were no good ones. Googling, too was of no avail.

Can someone please provide a simple explanation of this theorem (what exactly does it intend to convey) along with it's explanatory proof?

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    $\begingroup$ I edited the first equation in your question: you had written $$a_1x+b_1x+c_1=0$$ ... I suspect you meant to write $$a_1x+b_1y+c_1=0$$ $\endgroup$ – Zubin Mukerjee Aug 11 '17 at 14:13
  • $\begingroup$ @Abcd In case it helps explain Henning's (+1) answer, think of the special case where the lines have equation $y = 0$ ($a_{1} = c_{1} = 0$, $b_{1} = 1$) and $x = 0$ ($a_{2} = 1$, $b_{2} = c_{2} = 0$), so they meet at the origin. The claim is, every line through the origin has equation $y + \lambda x = 0$ for some $\lambda$. (Technically, by the way, you need to allow "$\lambda = \infty$" as a parameter value, interpreting the resulting equation as defining the second line.) $\endgroup$ – Andrew D. Hwang Aug 11 '17 at 15:53
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The claim is, written out in more words:

Suppose you have two lines, $\ell$ and $m$ with equations $$\ell: a_1x+b_1y+c_1 = 0 \\ m: a_2x+b_2y+c_2 = 0$$ Suppose further that $\ell$ and $m$ intersect, and that $k$ is a line through their point of intersection. Then there exists a real number $\lambda$ such that $$ a_1x+b_1y+c_1+\lambda\cdot(a_2x+b_2y+c_2) = 0 $$ is an equation for the line $k$.

Beware that this claim is actually only true if $\ell$, $m$, and $k$ are three different lines. (It also happens to be true when $\ell$ and $k$ are the same line, namely by setting $\lambda=0$, but that's more by accident than by design, I think).

The long equation at the bottom can also be rearranged as $$ (a_1+\lambda a_2)x + (b_1+\lambda b_2)y + (c_1+\lambda c_2) = 0 $$ so you should at least be able to recognize that it's always an equation for some line.

Why is this true, then? Well, the way the line equations work is that $a_1x+b_1y+c_1$ is actually an expression for the distance between $(x,y)$ and $\ell$ -- though not necessarily measured in the same units as $x$ and $y$ are, and with a sign such that it is negative on one side of $\ell$ and positive on the other side.

Clearly requiring this distance to be $0$ is the same as requiring $(x,y)$ to be on $\ell$ which is why it works as an equation for the line.

When we write $$ \underbrace{a_1x+b_1y+c_1}_{x'}+\lambda\cdot\underbrace{(a_2x+b_2y+c_2)}_{y'} = 0 $$ what we're doing is in effect to create a new coordinate system where $\ell$ and $m$ are the coordinate axes, and in that coordinate system describe the line $$ x' + \lambda y' = 0 $$ We can see this describes a line through the origin of the new coordinate system (which is just the intersection of $\ell$ and $m$). And every line through this origin can be represented in this way, except for the $x'$ axis, which is $m$.

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    $\begingroup$ Where does $\lambda$ come from? $\endgroup$ – user342531 Aug 11 '17 at 14:35
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    $\begingroup$ @Abcd: The theorem just promises you that a suitable $\lambda$ will exist -- it doesn't explain how to find it. In practice, if you know a second point on $k$ you can find $\lambda$ by plugging its coordinates into the long equation and solving for $\lambda$. $\endgroup$ – hmakholm left over Monica Aug 11 '17 at 14:37
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    $\begingroup$ How did it make it's way into the equation? What about the theorem's proof? $\endgroup$ – user342531 Aug 11 '17 at 14:40
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    $\begingroup$ @Abcd: I don't understand what you're asking. The theorem comes to you with that equation in it, including $\lambda$. The equation is what the theorem's claim is about. $\endgroup$ – hmakholm left over Monica Aug 11 '17 at 14:41
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    $\begingroup$ Could you explain this part: "Well, the way the line equations work is that is actually an expression for the distance between (x,y)and l" ? $\endgroup$ – user342531 Aug 16 '17 at 3:26
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If you assign $\lambda$ some value, the equation has the form $ax+by+c=0$, which is clearly the equation of a(nother) line.

And if you plug the coordinates of the intersection in the LHS, you get $0$, because it is also $0$ for the LHS of the first two equations. Hence this line contains the intersection.

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    $\begingroup$ As I understand it, the question is whether every line that contains the intersection can be written in that form? edit: I think you have shown that every line that can be written in that form contains the intersection $\endgroup$ – Zubin Mukerjee Aug 11 '17 at 14:17
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Hint: You can change the frame and suppose that the origin is the intersection of the line, the $x$-axis is parallel to the direction of $a_1x+b_1y+c_1$ and $y$ axis is parallel to $a_2x_2+b_2y+c_2$.

In this case every line throught the origin is of the form $ax+\lambda y=0$. Then you change the frame again to the original frame and obtain the requested expression.

Remark that $a_1x+b_1y+c_1+\lambda(a_2x+b_2y+c_2)=0$ does contains the line $a_2x+b_2y+c_2$, always, so it is better to write $\mu(a_1x+b_1y+c_1)+\lambda(a_2x+b_2y+c_2)=0$

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That is not a real answer, as I do not understand your question still, but a small illustration could be useful.

Take the two equations $x+y=0 \ \Leftrightarrow \ a_1=1, b_1=1, c_1=0$ and $x+2y+3=0 \ \Leftrightarrow \ a_2=1, b_2=2, c_2=3$. Or, we can rewrite them as functions $y=-x$ and $y=-\frac{1}{2}x - \frac{3}{2}$.

They intersect at the point $(3,-3)$. Your theorem says, we can find all the lines going through that point using the formula $(x+y) + \lambda (x+2y+3)=0$.

So, let's try this out.

$ x+y + \lambda x + 2\lambda y + 3\lambda=0 \ \Leftrightarrow \ y=\frac{1+\lambda}{-1-2\lambda} x + \frac{3\lambda}{-1-2\lambda}$, if we rewrite this as a function.

Put $x=3$, then $y(3)=\frac{3+3\lambda+3\lambda}{-1-2\lambda}=\frac{3(1+2\lambda)}{-1(1+2\lambda)}=-3$. That means we have found a family of lines, that all go through the point $(3,-3)$, as we see that $y(3)$ is independent from $\lambda$.

Writing this idea in common (not for $x+y$ and $x+2y+3$ in particular) will bring you a proof. You can find the intersection point $\left((\frac{-c_1-c_2}{a_1-a_2}),(\frac{-a_1c_2-c_1a_2}{b_1a_1-b_1a_2}), \ a_1-a_2\ne0, \ b_1a_1-b_1a_2 \ne 0\right)$ and put in the equation $y=\frac{-a_1-\lambda a_2}{b_1+\lambda b_2} \cdot x + \frac{-c_1-\lambda c_2}{b_1 + \lambda b_2}$, which you get, if you transform your given equation from the theorem.

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