In this lesson, I’ll cover a few examples to illustrate the concept of Family of Lines.

**Q1. **Find the equation of the line which passes through the point of intersection of L_{1}:2x – 3y + 5 = 0 and L_{2}: 3x – 2y + 5 = 0, and also passes through the origin.

**Sol. **Here’s one way to solve the problem.

(i) Find the point of intersection L_{1} and L_{2}: This point will be (-1, 1)

(ii) Use the two point form to find the required equation: (y – 0)=[(1-0)/(-1-0)](x – 0) or **x + y = 0**.

But that’s a boring method. I’m interested in family of lines. So here’s another method.

As discussed before, equation of any line passing through the point of intersection of the lines L_{1}=0 and L_{2} =0 will be of the form **L _{1}+λL_{2} =0**, where λ is a parameter.

That is, the required line will be of the form (2x – 3y + 5)+λ(3x – 2y + 5) = 0. That’s half the work done! But what about λ?

Another condition is given to us: The line passes through the origin as well. We can therefore substitute the coordinates to obtain the value of λ.

We have, 2(0)-3(0)+5 + λ(3(0)-2(0)+5) = 0, or λ=-1. And we’re done!

The required line is (2x – 3y + 5)+(-1)(3x – 2y + 5) = 0, or **x + y = 0**. We didn’t even have to find the point of intersection of the given lines. Saved some effort.

So remember, any line passing through the point of intersection of two given lines will be given by **L _{1}+λL_{2} =0**, and the value of the parameter λ will be determined by another given condition.

**Q2. **Show that, for all real values of ‘a’, the line ax + 3y + a – 6 = 0 passed through a fixed point. Also find the coordinates of that point.

**Sol. **Hmm.. what about this one?

Now, whenever you’re asked to prove that a line passes through a fixed point, think of family of lines.

The given line can be written as 3y – 6 + a(x + 1) = 0. This looks familiar.. of the form L_{1}+λL_{2} =0, where L_{1} is 3y – 6 and L_{2 }is x + 1, and a is the parameter λ.

Well? This represents a family of lines passing through the point of intersection of L_{1 }= 0 and L_{2 }= 0, or 3y – 6 =0, and x + 1 = 0, which is **(-1, 2)**.

Therefore, for all real values of a, the line ax + 3y + a – 6 passes through the (fixed) point (-1, 2). You may substitute the coordinates and verify this yourself.

**Q3. **If a + b + c = 0, then show that the line ax + by + c = 0 will always pass through a fixed point. Find the coordinates of that point.

**Sol. **This problem is quite similar to the previous one.

Using the given condition, the line can be written as ax + by – a – b = 0 or a(x – 1)+b(y – 1) = 0 or (x – 1) + (b/a)(y – 1) = 0.

Again, something of the form L_{1} + λL_{2} = 0. Therefore, the given line will pass through the point of intersection of the lines x – 1 = 0 and y – 1 = 0, which is **(1, 1)**.

I’ll stop here. Hope you’ve got the idea of the application of family of lines.

Head over to the next lesson, where I’ll be talking about angle bisectors.