Maths on the Line

We use the equation of the line – y=mx+c – for many things in geometry. It is one of those rather useful equations and well worth memorising and knowing how to use. Below we will look at two fundamental uses of it; finding C – that is where the line intersects the y axis, and identifying if a given point ∈ {|AB|} – that is the set of all possible points on a given line (in this case |AB|).


Identifying C

Every point on an x-y axis has a set of coordinates, an x and a y value. Here we see that point A has an x value of -5 and a y value of -1. Point B likewise has x and y values – of 3 and 4 respectively. Since we know that point C is where the line crosses the y axis we know that the value of y at C is 0 – by definition. So when we are looking for C we are really only looking for the x value of C.

So, what do we know?

We know that A is -5,-1 and that B is 3,4. We also know that y at C is 0. Then we have our equation of the line: y=mx+c. This gives us our first problem. We don’t know m. This m refers to the gradient or the ‘slope’ of the line, and we have to find the value of it for this equation.

m = Δy/Δx
This looks a bit off-putting, but there is nothing to worry about. This Δ – or delta – only means the difference in the value of y at points A and B over the difference in the value of x at points A and B. Written like this:

m = (y2-y1)/(x2-x1)
Point A = x1y1. and Point B = x2y2.
This would make our equation:
m = (4- -1)/(3- -5) ∨ (5)/(8)
m = 5/8 ⇒ 0.625

This means that for every unit the value of x increases the value of y increases by 0.625, and for every unit decrease of x the value of y decreases by 0.625.

Now we can return to our equation y=mx+c and fill in the blanks. We can use the x and y coordinates of either A or B, but not a mixture of the two. We will get the same result no matter which one we use.

For A:
-1 = 0.625(-5) + c
-1 = -3.125 + c
3.125 -1 = c
c = 2.125

For B:
4 = 0.625(3) + c

4 = 1.875 + c
4 -1.875 = c
c = 2.125

Proving a Point is on the Line

Supposing we were told of another point (P). We are asked to show that P (11,9) is on |AB|. This is easy, because we have already done most of the work. All that we are required to do is substitute the x and y coordinates of P into our our y=mx+c equation, remembering that m = 0.625 and c = 2.125.

For P:
9 = 0.625(11) + 2.125
9 = 9
∴ P belongs to |AB|

Supposing then we are given another point (Q), and we are asked to find whether Q(-13,-5) belongs to |AB| – we simply do the same again:

For Q:
-5 = 0.625(-13) + 2.125
-5 ≠ -6
∴ Q does not belong to |AB|

Here we have shown that Q is not ∈ {|AB|}. Using this formula we can discover both if points belong to or do not belong to the line in question. So these are the basics of the equation of the line. I hope this was of some help to you, and – as always – if you have any comments or questions please post them in the Thoughts and Questions section below.


Thoughts and Questions

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