Intersection and Concurrency

This lesson will talk about intersection of two lines, and concurrency of three lines.

To be precise, we’re dealing with two questions here:

1) How do we find out the point of intersection of two lines?

2) How can we tell whether 3 lines are concurrent (i.e. pass through the same point)?

The first one is quite simple.

Suppose the lines 2x + y = 4 and x + y = 3 intersect at the point (x’, y’).

That means, (x’, y’) must lie both the lines, or its coordinates must satisfy both the equations.

We can write the relations 2x’ + y’ = 4 and x’ + y’ = 3, and solving these for x’ and y’, we get x’ = 1 and y’ = 2.

Therefore the point of intersection will be (1, 2).

We can ignore the x’ and y’, and directly solve the two equations, 2x + y =4 and x + y = 3. The solution of these two equations will give the coordinates of the point of intersection. As simple as that.

There may be some cases in which we will not get a solution. Here’s an example.

Consider the lines 2x + y = 4 and 4x + 2y = 0. Try and solve these two equations. You won’t get any solution !

What does that mean geometrically? Well it means that the lines do not intersect. Or, the lines must be parallel.

Here’s a figure to illustrate both the cases..

The method of solving equations will give the points of intersection of any two curves in general.

For example, if we want to find out the intersection of this strange looking curve x2 + 2y = 3x and the line x + y = 1, we just have to solve these equations to get the values of x and y.

If we get more than one solution, that means, there is more than one intersection point. Similarly, no solution implies the curves do not intersect.

To the second question. What if 3 lines are concurrent?

That simply means one of them must pass through the point of intersection of the other two.

I’ll write this in a more formal way.

Let the equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 represent three different lines.

If they’re concurrent, then the point of intersection of the first two (or any two) lines must lie on the third.

This is quite straightforward. The point of intersection of the first two lines will be: $$(\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}, \frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1})$$

Before moving on, we need to take care about that term in the denominator, i.e. a1b2-a2b1. What if it equals zero?

Well, it cannot. Why? Because if it does, then a1b2-a2b1 = 0, will lead to -a1 / b1 = -a2 / b2 , which implies the slope of the lines will be equal, which means the lines will be parallel, which means that they cannot intersect ! (I gave a hint about slopes here)

But they do ! Therefore, a1b– a2b≠ 0. I’ll come back to this later.

Moving on. Now, for concurrency, the third line must also pass through this point. Therefore, the point’s coordinates must satisfy the line’s equation.

We’ll finally obtain the relation: a3(b1c2 – b2c1) + b3(c1a2 – c2a1) + c3(a1b2 – a2b1) = 0

This relation can be written in a fancy way as: $$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\\\end{vmatrix}=0$$

That awesome looking 3×3 arrangement of things is known as a determinant. And it is easier to remember than the previous scary result.

Don’t worry if you don’t know what that means. I’ll discuss that a bit later (much later actually, so search for it on Wikipedia or something, if you’re curious.)

I’m done here. See you later. Goodbye.