This lesson will cover a few examples involving the two-point form of the equation of a straight line.
Example 1 Find the equation of the line passing through (2, 4) and (4, 2).
Solution We have the two points, and we know the two-point form of the equation. Let’s substitute the given coordiantes in the equation. We’ll get:
y – 4 = {(2 – 4) / (4 – 2)} (x – 2)
or
x + y – 6 = 0
This is what the line looks like.
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’s 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!
To do that, we’ll find 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 = {(4 – 2) / (3 – 1)} (x – 1)
This simplies to:
x – y + 1 = 0
To find if C lies on this line, we’ll substitute its coordinates in the previous equation, and check if LHS = RHS.
On substituting, we’ll get
LHS = (-2) – (-1) + 1 = 0 = RHS
This means that C lies on the line x – y + 1 = 0, which passes through A and B.
This means A, B, and C all lie on the same line, or they are collinear! Here’s a figure for illustration.
Example 3 A triangle has the vertices as A(0, 0), B(3, 0) and C(0, 4). Find the equation of
(i) the median through A
(ii) the angle bisector through A
Solution (i) We know that the median passes through a vertex and the mid point of the side opposite to it.
We already know the coordinates of the vertex A, i.e. (0, 0). Let’s find the midpoint D of BC. Then, we’ll use the two point form to find the equation of the median AD.
The mid-point of B and C is:
((3 + 0)/2, (0 + 4)/2) or (3/2, 2)
We have the two points. Let’s find the equation now. Using the two point form, we’ll get:
y – 0 = {(2 – 0) / (3/2 – 0)} (x – 0)
This simplifies to:
4x – 3y = 0
(ii) In this case, the angle bisector passes through the vertex and the incenter of the triangle.
The coordinates of the incenter can be found easily. I’ll skip the calculations here. Using the formula decribed here, the incenter of the triangle is (1, 1).
We have the two points again, i.e. the vertex (0, 0) and the incenter (1, 1). Using the two point form, the equation of the angle bisector will be:
y – 0 = {(1 – 0) / (1 – 0)} (x – 0)
This simplifies to:
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.