# 4.2 Co-ordinate geometry

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## 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 $\left({x}_{1};{y}_{1}\right)$ and $\left({x}_{2};{y}_{2}\right)$ , and we know that the general form of the equation for a straight line is:

$y=mx+c$

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

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

where $\left({x}_{1};{y}_{1}\right)$ 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:

$\begin{array}{ccc}\hfill {y}_{1}& =& m{x}_{1}+c\hfill \\ \hfill {y}_{2}& =& m{x}_{2}+c\hfill \end{array}$

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

$\begin{array}{ccc}\hfill {\mathrm{y}}_{2}-{\mathrm{y}}_{1}& =& m{x}_{2}-m{x}_{1}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ \hfill {\mathrm{y}}_{1}& =& m{x}_{1}+c\hfill \\ \hfill c& =& {y}_{1}-m{x}_{1}\hfill \end{array}$

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

$\begin{array}{ccc}\hfill y& =& mx+c\hfill \\ & =& mx+\left({y}_{1}-m{x}_{1}\right)\hfill \\ \hfill \mathrm{y}-{\mathrm{y}}_{1}& =& m\left(x-{x}_{1}\right)\hfill \end{array}$

If you are asked to calculate the equation of a line passing through two points, use:

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

to calculate $m$ and then use:

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

to determine the equation.

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

$\begin{array}{ccc}\hfill m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ & =& \frac{2-1}{2-\left(-1\right)}\hfill \\ & =& \frac{1}{3}\hfill \end{array}$

and then substituting this value into

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

to obtain

$\begin{array}{ccc}\hfill y-{y}_{1}& =& \frac{1}{3}\left(x-{x}_{1}\right).\hfill \end{array}$

Then substitute $\left(-1;1\right)$ to obtain

$\begin{array}{ccc}\hfill y-\left(1\right)& =& \frac{1}{3}\left(x-\left(-1\right)\right)\hfill \\ \hfill y-1& =& \frac{1}{3}x+\frac{1}{3}\hfill \\ \hfill y& =& \frac{1}{3}x+\frac{1}{3}+1\hfill \\ \hfill y& =& \frac{1}{3}x+\frac{4}{3}\hfill \end{array}$

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

Find the equation of the straight line passing through $\left(-3;2\right)$ and $\left(5;8\right)$ .

1. $\begin{array}{ccc}\hfill \left({x}_{1};{y}_{1}\right)& =& \left(-3;2\right)\hfill \\ \hfill \left({x}_{2};{y}_{2}\right)& =& \left(5;8\right)\hfill \end{array}$
2. $\begin{array}{ccc}\hfill m& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ & =& \frac{8-2}{5-\left(-3\right)}\hfill \\ & =& \frac{6}{5+3}\hfill \\ & =& \frac{6}{8}\hfill \\ & =& \frac{3}{4}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-\left(2\right)& =& \frac{3}{4}\left(x-\left(-3\right)\right)\hfill \\ \hfill y& =& \frac{3}{4}\left(x+3\right)+2\hfill \\ & =& \frac{3}{4}x+\frac{3}{4}·3+2\hfill \\ & =& \frac{3}{4}x+\frac{9}{4}+\frac{8}{4}\hfill \\ & =& \frac{3}{4}x+\frac{17}{4}\hfill \end{array}$
4. The equation of the straight line that passes through $\left(-3;2\right)$ and $\left(5;8\right)$ is $y=\frac{3}{4}x+\frac{17}{4}$ .

## 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, $\left({x}_{1};{y}_{1}\right)$ , 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:

$\begin{array}{ccc}\hfill \mathrm{If the lines are parallel, then}\phantom{\rule{1.em}{0ex}}\mathrm{m}& =& {m}_{0}\hfill \\ \hfill \mathrm{If the lines are perpendicular, then}\phantom{\rule{1.em}{0ex}}\mathrm{m}×{\mathrm{m}}_{0}& =& -1\hfill \end{array}$

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

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

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 $\left(-1;1\right)$ .

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

$m=2$

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

$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-1& =& 2\left(x-\left(-1\right)\hfill \\ \hfill y-1& =& 2\left(x+1\right)\hfill \\ \hfill y-1& =& 2x+2\hfill \\ \hfill y& =& 2x+2+1\hfill \\ \hfill y& =& 2x+3\hfill \end{array}$ The equation of the line passing through ( - 1 ; 1 ) and parallel to y = 2 x - 1 is y = 2 x + 3 . It can be seen that the lines are parallel to each other. You can test this by using your ruler and measuring the perpendicular distance between the lines at different points.

## Inclination of a line (a) A line makes an angle θ with the x -axis. (b) The angle is dependent on the gradient. If the gradient of f is m f and the gradient of g is m g then m f > m g and θ f > θ g .

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.

$m=\frac{\Delta y}{\Delta x}$

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

$\begin{array}{ccc}\hfill tan\theta & =& \frac{\Delta y}{\Delta x}\hfill \\ \hfill \therefore m& =& tan\theta \hfill \end{array}$

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

$\begin{array}{ccc}\hfill \therefore tan\theta & =& 1\hfill \\ \hfill \therefore \theta & =& {45}^{\circ }\hfill \end{array}$

## Co-ordinate geometry

1. Find the equations of the following lines
1. through points $\left(-1;3\right)$ and $\left(1;4\right)$
2. through points $\left(7;-3\right)$ and $\left(0;4\right)$
3. parallel to $y=\frac{1}{2}x+3$ passing through $\left(-1;3\right)$
4. perpendicular to $y=-\frac{1}{2}x+3$ passing through $\left(-1;2\right)$
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 ${90}^{\circ }$ .)

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