This lesson will cover a few examples involving the two-point form of the equation of a straight line.

I’ll start with an easy one.

**Example 1 **Find the equation of the line passing through (2,4) and (4,2).

**Solution **No need for much explanation here. We have the two points, and we know the two-point form of the equation. Therefore the required equation is y – 4 = \(\frac{(2 – 4)}{(4-2)}\) ( x – 2) or **x + y – 6 = 0**

Boring.. next..

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

**Solution **We already know three methods (explained here, here and here) of proving collinearity. Here is yet another one !

If we’re able to show that one of the point lies on the line joining the other two, then we’re done! How can we do this?

Find out the equation of the line joining any two of the given points, and show that the coordinates of the third also satisfy that equation.

Now, the equation of the line joining A and B is y – 2 = \(\frac{(4-2)}{(3-1)}\) (x – 1) or **x – y + 1 = 0.**

Clearly, we can see that the coordinates of C also satisfy this equation: (-2) – (-1) + 1 = 0.

Which means that C lies on the line x – y + 1 = 0. But this is the line which passes through A and B. This means A, B, and C all lie on the same line, or they are collinear !

Easy enough ! Next one..

**Example 3 **Find the equation of (i) the median and (ii) the angle bisector, through the vertex A of the triangle which has the vertices as A(0, 0), B(3, 0) and C(0, 4)

**Solution **(i) Notice the two conditions, one being passing through a point (0,0) and the other that it is a special line – the median.

How to use the second condition?

Simple ! The median passes through the mid-point of the opposite side (i.e. BC), which is (3/2, 2).

So, now we have two points through which the line will pass, and we’ll use the two-point form of the equation.

The required equation is y – 0 = \( \frac{(2 – 0)}{(3/2 – 0)} \) (x – 0) or **4x – 3y = 0**

(ii) What about this one? You guessed it right ! The angle bisector will also pass through the incenter of the triangle which will be (1, 1) (I’ve skipped all the calculations. Go back here for the formulas)

Two points again ! The required equation is given by y – 0 = \(\frac{(1-0)}{(1 – 0)}\) (x – 0) or **x – y = 0**

Alright, that’s it for now. See you in the next lesson, where I’ll talk about another form of the equation, known as the normal form.