4.2 Co-ordinate geometry

Co-ordinate geometry

Equation of a line between two points

There are many different methods of specifying the requirements for determining the equation of a straight line. One option is to find the equation of a straight line, when two points are given.

Assume that the two points are $(x_1;y_1)$ and $(x_2;y_2)$ , and we know that the general form of the equation for a straight line is:

So, to determine the equation of the line passing through our two points, we need to determine values for $m$ (the gradient of the line) and $c$ (the $y$ -intercept of the line). The resulting equation is

where $(x_1;y_1)$ are the co-ordinates of either given point.

Finding the second equation for a straight line

This is an example of a set of simultaneous equations, because we can write:

We now have two equations, with two unknowns, $m$ and $c$ .

Now, to make things a bit easier to remember, substitute [link] into [link] :

For example, the equation of the straight line passing through $(-1;1)$ and $(2;2)$ is given by first calculating $m$

and then substituting this value into

to obtain

Then substitute $(-1;1)$ to obtain

So, $y=\frac{1}{3}x+\frac{4}{3}$ passes through $(-1;1)$ and $(2;2)$ .

Equation of a line through one point and parallel or perpendicular to another line

Another method of determining the equation of a straight-line is to be given one point, $(x_1;y_1)$ , and to be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is $y=mx+c$ and the equation of the second line is $y=m_{0}x+c_0$ , then we know the following:

Once we have determined a value for $m$ , we can then use the given point together with:

to determine the equation of the line.

For example, find the equation of the line that is parallel to $y=2x-1$ and that passes through $(-1;1)$ .

First we determine $m$ , the slope of the line we are trying to find. Since the line we are looking for is parallel to $y=2x-1$ ,

The equation is found by substituting $m$ and $(-1;1)$ into:

Inclination of a line

In [link] (a), we see that the line makes an angle $\theta$ with the $x$ -axis. This angle is known as the inclination of the line and it is sometimes interesting to know what the value of $\theta$ is.

Firstly, we note that if the gradient changes, then the value of $\theta$ changes ( [link] (b)), so we suspect that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in the $y$ -direction to a change in the $x$ -direction.

But, in [link] (a) we see that

For example, to find the inclination of the line $y=x$ , we know $m=1$

Co-ordinate geometry

1. Find the equations of the following lines
   1. through points $(-1;3)$ and $(1;4)$
   2. through points $(7;-3)$ and $(0;4)$
   3. parallel to $y=\frac{1}{2}x+3$ passing through $(-1;3)$
   4. perpendicular to $y=-\frac{1}{2}x+3$ passing through $(-1;2)$
   5. perpendicular to $2y+x=6$ passing through the origin
2. Find the inclination of the following lines
   1. $y=2x-3$
   2. $y=\frac{1}{3}x-7$
   3. $4y=3x+8$
   4. $y=-\frac{2}{3}x+3$ (Hint: if $m$ is negative $\theta$ must be in the second quadrant)
   5. $3y+x-3=0$
3. Show that the line $y=k$ for any constant $k$ is parallel to the x-axis. (Hint: Show that the inclination of this line is $0^{\circ }$ .)
4. Show that the line $x=k$ for any constant $k$ is parallel to the y-axis. (Hint: Show that the inclination of this line is ${90}^{\circ }$ .)