## Family of Lines

Today, I’ll discuss quite an interesting topic in straight lines – **Family of Lines**.

Consider two lines, L_{1}: 2x + 3y – 5 = 0 and L_{2}: 2x – y – 1 = 0. These lines intersect at the point (1, 1). (obtained by solving the two equations).

Now consider the line obtained by adding the equations of L_{1} and L_{2}, i.e. 4x + 2y – 6 = 0 (a different line).

Note that this line also passes through (1,1). A coincidence? Maybe..

Let’s take another line, obtained by adding L_{1} and twice of L_{2}, i.e. (2x + 3y – 5) + 2(2x – y – 1) = 0, or 6x + y – 7 = 0 (another different line).

Again, this line passes through the point (1, 1). Hmm.. doesn’t look lie a coincidence now. Or does it?

One more example. I’ll take three times the first equation and subtract the second from it. We get 4x + 10y – 14 = 0, which again passes through the point (1, 1). What’s going on?

Here’s what’s happening. Consider a line of the form a(L_{1}) + b(L_{2}) = 0, i.e. a(2x + 3y – 5)+b(2x – y – 1) = 0 (where a and b are real numbers, not both 0).

For different values of a and b, you’ll get different lines, as you’ve seen in the above examples. And all of these lines will pass through point (1,1) (which is the point of intersection of the two lines)

Why? Because, if you substitute the coordinates (1,1) in the above equation, the two terms a**(2x + 3y – 5)** and b**(2x – y – 1)** will separately become zero, (as (1,1) lies on both of these lines) and thus making LHS zero (implying the point (1,1) lies on the line).

That explains it all ! Now if you put any value of a and b, for example a = 1 and b = 1 (the first example above), we’ll get 1(2x + 3y – 5)+1(2x – y – 1) = 0 or 6x + y – 7 = 0, which will pass through (1,1), because of the reason explained just now.

Awesome!

Therefore, the equation a(2x + 3y – 5)+b(2x – y – 1) = 0 represents a ‘family’ of lines, all members of which pass through the point of intersection of 2x + 3y – 5 = 0 and 2x – y – 1 = 0 (as shown in the figure above).

This can also be written as (2x + 3y – 5) + (b/a)(2x – y – 1) = 0 (assuming a to be non-zero) or **(2x + 3y – 5) + λ(2x – y – 1) = 0**, where λ is a parameter, whose different values will give different lines, all passing through (1,1).

## Lesson Summary

**L**, where λ is a parameter, represents a family of lines, all of which pass through the point of intersection of L_{1}+ λL_{2}= 0_{1}=0 and L_{2}=0.- Any line passing through the point of intersection of L
_{1}=0 and L_{2}=0, will have its equation of the form**L**, where λ is a parameter._{1}+ λL_{2}= 0