Concepts involved in "Family of Lines". 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? 
 A: 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 $m$.
A: 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.
A: 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 $\mu(a_1x+b_1y+c_1)+\lambda(a_2x+b_2y+c_2)=0$
A: 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.
