6. Section Formula: Examples

Welcome ! In this lesson I’ll cover some simple examples involving the section formula.

Example 1 Find the coordinates of the point which divides AB, where A and B have the coordinates (2,4) and (4,6) respectively, in the ratio 1:3 (i) internally (ii) externally

Solution The figure illustrates the two cases. Notice that in both cases AC:CB=1:3

Section Formula examples

Fig. 1: The two cases

We just have to put in the values in the section formula

(i) x\(=\frac{1\times4+3\times2}{1+3}=\frac{5}{2}\) and y\(=\frac{1\times6+3\times4}{1+3}=\frac{9}{2}\) Therefore the coordinates of the required point are \((\frac{5}{2},\frac{9}{2})\)

(ii) x=\(\frac{1\times4-3\times2}{1-3}=\)1 and y\(=\frac{1\times6-3\times4}{1-3}=\)3 Therefore the coordinates of the required point are (1,3)

The only thing you need to take care of is the order of A and B. The answer would have been different if the question had mentioned BA instead of AB. Always make a figure to avoid any errors.

On to the next..

Example 2 Find the ratio in which the point C(2, 4) divides the line segment AB where A and B have the coordinates as (i) (0, 0) and (4, 8) respectively  (ii) (0, 0) and (1, 2) respectively

Solution This problem is the reverse of the previous one. We have been given the three points, and have to find out the ratio in which one divides the line joining the other two.

(i) Let the ratio be k:1 (we can also assume it to be m:n, but this is equal to m/n:1, and m/n can be taken equal to some number k. So it doesn’t make a difference.) Now, applying the section formula we can write 2=\(\frac{k\times4+1\times0}{k+1}\), which on solving gives k=1. Therefore it divides AB in the ratio is 1:1 (making it the midpoint of AB) Wait.. but what about the y-coordinate? Lets check using that too. 4\(=\frac{k\times8+1\times0}{k+1}\). Again we have k=1.

Hmm.. somethings strange! How did both equations magically gave the same answer? Do they have to? I don’t know.. Lets change the point (4, 8) to (4,9). If we calculate the value of k using the y coordinate of C now, we get k=4/5, or the ratio as 4:5. That is the same point divides the same line in two different ratios! That’s not possible.

So what’s wrong then? When B is changed to (4, 9). The three points are not collinear any more. The point C doesn’t even lie on the line joining A and B, forget about dividing it in ratios, that’s why the silly result !

So, in case the two equations (both of which you must solve) give different values of k, it means the points are non-collinear. (That gives us an alternate way to check whether three points are collinear or not. I’ll explain this in a subsequent example.) Lets go to the next part.

(ii) Again, lets begin with assuming the ratio to be k:1 Putting in the values, we have 2=\(\frac{k\times1+1\times0}{k+1}\) which gives k=-2  (Also if we check with the y-coordinate of C we have 4=\(\frac{k\times2+1\times0}{k+1}\), which again gives k=-2)

What do you think the negative value of k implies? It means that the point C divides AB in the ratio 2:1 externally.

Since we’ve assumed in out calculation the division to be internal, the negative value shows that it was actually the opposite. In case we had used the formula for external division, we would have got the value of k as 2. Here is a figure to illustrate what’s happening

Section Formula examples

Fig. 2: Relative positions of A, B and C

Next !

Example 3 Find the distance between the incentre and the centroid of the triangle whose vertices are (0, 0), (3, 0) and (0, 4)

Solution This is an easy problem. We’ll first find the centroid, followed by the incentre and calculate the required distance using the distance formula. The centroid is \((\frac{0+3+0}{3},\frac{0+0+4}{3})\), which is (1, 4/3)

The sides of the triangle are of lengths 3, 4 and 5 (calculated using distance formula). Therefore the incentre is given by \((\frac{5\times0+4\times3+3\times0}{5+4+3},  \frac{5\times0+4\times0+3\times4}{5+4+3})\) which is (1, 1)

The required distance is given by \(\sqrt{(1-1)^2+(4/3-1)^2}\)= 1/3

Example 4 Prove that the points A(1,2), B(3,10) and C(2,6) are collinear.

Solution We’ll use the section formula to prove collinearity instead of the distance formula. Remember that three points will be collinear if we’re able to prove that one divides the line joining the other two in some ratio (any ratio).

We’ll assume the ratio in which B divides AC to be k:1 and calculate the values of k using both x and y coordinates of B. If they’re same, then we’re done ! The first equation is 3=\(\frac{k\times2+1\times1}{k+1}\). This gives k=-2 (that is, 2:1 externally) The second equation is 10\(=\frac{k\times6+1\times2}{k+1}\). This also gives k=-2.

That means B divides AC in the ratio 2:1 externally, which is only possible if they are collinear.

Example 5 Show that the points A(0, 0), B(3, 0), C(4, 1) and D(1, 1) form a parallelogram

Solution This the same question as 3(i) in lesson 3. But we’ll solve it using a much better method. A sufficient condition for ABCD to be a parallelogram is that the diagonal AC and BD bisect each other. (Hint: congruent triangles).

This means that AC and BD have the same midpoint. So, all we have to do is that prove that AC and BD have the same midpoints. We can see that it is actually true. The midpoint of both AC and BD is (2, 1/2), making ABCD a parallelogram.

That’s it for now. A long way to go. Hope you’re enjoying. See you in the next lesson!

Leave a comment