# straight Lines

You will be expected to be able to find the equation of a straight line given a variety of scenarios. This could be as simple as finding the equation of a line given two points, or it could be finding the equation of a parallel or perpendicular line given points or other relevant information. You may also be expected to represent straight lines in different formats including the following:

## Line Equations given Two Points

Recall that to find the equation of a straight line between two points, you must first find the gradient of the line that connects the two points. We can calculate this gradient using rise over run. One of the points can then be used to calculate the y-intercept of the line. Remind yourself of how to do this with the following example.

Example – Find the equation of the line between the points (2,4) and (4,10).

It is always a good idea to sketch the information first – sometimes this helps you realise what to do next if you get stuck. The line that connects these two points rises by 6 as it runs along 2 and so rise over run is . It follows that the gradient of the line is 3. Hence, this tells us that the line is of the form . To then find , plug one of the given points into the equation so far: and so c must be -2. The final equation of the line is .

## Parallel and Perpendicular Lines

You may be required to find the equation of a straight line that is either parallel or perpendicular to another given line. This type of question varies in its complexity. For instance, you could be given some very basic information as in Example 1 below or it can be much more involved as in Example 2.

Note that parallel lines have the same gradient – see Example 1. However, if the gradient of a given line is then the gradient of a line perpendicular to it is – see Examples 1 and 2.

When studying differentiation, you will meet tangents and normals – these are examples of perpendicular lines.

## Proportionality

Consider the straight lines that intersect the -axis at the origin. For this type of straight line the y-intercept is 0 in the straight line equation and so they are of the form . We tend to use instead of when we talk about the gradient of a line that goes through the origin because is known as the constant of proportionality. When the straight line takes an equation of this form, we say that is proportional to or . Click here to see more on proportionality.

Example – Recall that the circumference of a circle () is found by multiplying the diameter of the circle () by . The circumference of a circle is thus proportional to the diameter and the constant of proportionality is – we write this as . Zero diameter means zero circumference and so the plot of circumference against diameter goes through the origin.

This isn’t quite as simple as the plot of , for example, but is still a constant all the same. Note that circumference is also proportional to radius but the constant of proportionality is then .

## Modelling with Straight Lines

This table shows some temperatures given in degrees Celsius and their corresponding temperatures in degrees Fahrenheit.

From the graph below we can see that the relationship between Celsius and Fahrenheit is precisely a linear one. More specifically, the gradient is and the y-intercept is and so :

This equation can now be used to convert any temperature given in Celsius to Fahrenheit and vice versa.

## Examples

1. What is the equation of the line that runs parallel to and intersects the y-axis at ?
2. Find the equation of the line that is perpendicular to the line and intersects it at the point (2,7). Give your answer in the form where , and are integers.

Solution:

1. Parallel lines have the same gradient and so the line required has gradient 3. Since it intersects the -axis at (0,4), the y-intercept is 4 and so the equation is given by .

Always draw sketches to accompany your working – they help with problem solving and allow you to check that your answers make sense.

2. The line has gradient 2 and so a line perpendicular to it has gradient . The equation so far is . We can find by inserting the coordinates giving . Hence, and the equation of the line is . Multiplying through by 2 to make all coefficients integer gives . Finally, we have the equation in the required form: .

The line has equation . The line is perpendicular to , intersecting it at the point , and also crosses the -axis at the point . Find the distance and the area of the triangle .

Solution:

We start by sketching the information as this will tell us more about the problem. To find the distance we require the coordinates of point . To do this, we must first find the equation of line . The line can be rewritten as , which has gradient . The gradient of is thus -3 and if it intersects the -axis at 5, it must intersect the -axis at 15, using rise over run. Hence, has equation . To find the coordinates of , the intersection of the two lines, we solve the simultaneous equations: and , to get and

The length can now be found using Pythagoras: to 2 decimal places. The area of triangle can now also be found using the height of the triangle which is 3.9 and the length of its base which is 5, i.e. .

## Videos

A past exam question (and solutions) testing knowledge and problem solving in coordinate geometry: straight lines.

Coordinate Geometry past AS exam question and solutions – straight lines and GCSE trigonometry.

Tricky coordinate geometry past AS Maths exam question – tests knowledge of straight lines, problem solving and also brings in a quadrilateral at the end.