4. Section Formula

In this lesson we’ll establish the formula to find out the coordinates of a point, which divides the line segment joining two given points in a given ratio. The formula is known as the Section Formula. Let us begin !

Section Formula

Consider two points P(x1, y1) and Q(x2, y2). We have to find the coordinates of the point R which divides PQ in the ratio m:n, i.e. PR/RQ=m/n.

There can be two cases, R lying between P and Q, or outside the line segment PQ. The following figure gives some idea about what I’m saying.

Section formula internal and external division

Fig. 1: The two cases

(Notice that in the above figure ‘m’ and ‘n’ do not denote the lengths of PR and QR. They are just indicating the proportion.)

Let us consider both these cases.

Case I – R lying between P and Q

I’ll first illustrate an ugly (but valid) method to compute the coordinates of R, by using what we already know – the distance formula. As I mentioned earlier that there are (in general) multiple methods to solve a coordinate geometry problem. This is one of those cases. The idea is to develop your thinking level, and give you an idea of which methods are good and which aren’t.

Here’s one way you can think: “Since there are two unknowns ‘x’ and ‘y’ (i.e. the coordinates of R), I need to find two equations, or two geometric conditions, which I’ll convert into (solvable) equations, using the formulas I know.” (Again, we’re dealing with geometric conditions using algebra)

Hmm.. one condition is already given, PR / RQ = m / n, or n.PR = m.RQ. So we can write \(n\sqrt{(x-x_1)^2+(y-y_1)^2}=m\sqrt{(x_2-x)^2+(y_2-y)^2}\)

Its already starting to get ugly.

What about the next condition? The first condition does not ensure that P, R and Q will lie on the same line. So, we have our second condition, PR + RQ = PQ. We get an even more ugly equation –


I’ll stop here. You might have got the idea why this is not the best method. In fact, as we’ll see later on, methods involving distance formula generally get quite complicated and difficult to solve.

So what is our alternative? Time for some geometry, let’s make some constructions.

Section formula internal division

Fig. 2: Internal division

I’ve drawn RA & QB parallel to the Y axis, and PA & RB parallel to the X axis. The triangles RPA and QRB are similar, by AA similarity.

How does this help? We are given PR / QR, which, using similarity, is equal to RA / QB and PA / RB. Let me write this in a better way.

\(\frac{PR}{QR} = \frac{ RA}{QB} = \frac{PA}{RB} = \frac{m}{n}\)

Now, PA = x – x1 and RB = x2 – x (I’ve provided explanation for this in a previous lesson)

Therefore, we can write \(\frac{PA}{RB}=\frac{x-x_1}{x_2-x}=\frac{m}{n}\)

On solving the above for x, we get x=\(\frac{mx_2+nx_1}{m+n}\)

What about y? RA=y – y1 and QB=y2 – y, now repeat the above process to get y=\(\frac{my_2+ny_1}{m+n}\)

Therefore, we have the coordinates of the point R which divides PQ internally in the ratio m:n.

Internally? That’s the official term for case (i), when R lies somewhere between P and Q

Case II – R lying outside PQ

The given condition is the same, i.e. PR/RQ=m/n, but the figure will be a little different

Section formula external division

Fig. 3: External division

In this case R divides PQ externally in the ratio m:n.

I’m not going to derive this one. Please try to do it yourself. The method is the same as in the first case – find the similar triangles, express the given ratio (m/n) in terms of  x1 , x, y1 and y2.

If you do it right, you should get x=\( \large \frac{mx_2-nx_1}{m-n}\) and y=\( \large \frac{my_2-ny_1}{m-n}\)

Lesson Summary

  1. The coordinates of the point dividing the line segment joining (x1, y1) and (x2, y2in the ratio m : n internally is given by \(\ \large \left ( \frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} \right ) \)
  2. The coordinates of the point dividing the line segment joining (x1, y1) and (x2, y2in the ratio m : n externally is given by \(\ \large \left ( \frac{mx_2-nx_1}{m-n} , \frac{my_2+ny_1}{m+n} \right ) \)

Well.. that’s it ! See you in the next lesson with some examples and applications !

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