# 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 $y=mx+c$, $ax+by+c=0$, or $y-y_1=m(x-x_1)$. Recall that the equation of a straight line is $y=mx+c$ where $m$ is the gradient and $c$ is the $y$-intercept.

## Line Equations given Two Points

## Parallel and Perpendicular

You may be required to find the equation of a straight line that is either parallel or perpendicular to another given line. This could be directly where you asked to find an equation in no uncertain terms (see Example 2). Alternatively, the question may be stated indirectly, where you are asked for the area of a triangle, for example, which requires finding the equations first (see Example 3).

Note that **parallel lines have the same gradient**.

If the gradient of a given line is $m$ then the gradient of the line **perpendicular to it is $-\frac{1}{m}$. **See Examples 2 and 3.

Also note that tangents and normals are also straight lines that have these same properties. Tangents and Normals require differentiation to identify the gradient of the curve at a specific location. See Differentiation and Tangent & Normals.

## Proportionality

Consider the straight lines that cut the $y$-axis at the origin, i.e. $c=0$ in the straight line equation. Typically, these are equations of the form $y=kx$ where $k$ is a specified constant (or a constant to be found). We tend to use k instead of $m$ when we talk about the gradient of a line that goes through the origin. $k$ is the constant of proportionality. When the straight line takes an equation of this form, we say that $y$ is proportional to $x$ or $y\propto x$. See Example 4.

Click here to see some examples of proportionality in physics. Note that the constant of proportionality may also be negative. This just means that the straight line will have a negative gradient and still pass through the origin.

## Modelling with Straight Lines

A mathematical model is a representation of a real-life situation using mathematical concepts. For example, given a set of paired data, we might like to fit a straight line to it. See Correlation for more on this. Note that we are assuming that the paired data, when plotted, looks like a straight line (or roughly like a straight line). The fitted line, or the line of best fit when the data doesn’t fit exactly, can then be found to predict values of other unobserved pairs. See Example 5.

Click here to see another explanation of modelling with straight lines.

## Examples of Straight Lines

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

The line that connects these two points rises by 6 as it runs along 2 and so rise over run is 6/2 which equals 3. It follows that the gradient of the line is 3.

Hence, this tells us that the line is of the form $y=3x+c$. To then find $c$, plug one of the given points into the equation so far:

$4=3\times 2 +c$ and so c must be -2.

The final equation of the line as $y=3x-2$.

### (Parallel & Perpendicular)

- What is the equation of the line that runs parallel to $y=3x-1$ and intersects the $y$-axis at $(0,4)$?
- Find the equation of the line that is perpendicular to the line $y=2x+3$ and intersects it at the point (2,7). Give your answer in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers.

- Parallel lines have the same gradient and so the line required has gradient 3. Since it intersects the y-axis at (0,4), the y-intercept is 4 and so the equation is given by [math]y=3x+4[/math].
- The line perpendicular to $y=2x+3$ ($m=2$) has gradient $-\frac{1}{2}$ (taking $-1/m$). The equation so far is $y=-\frac{1}{2}x+c$. We can find $c$ by inserting the coordinates $(2,7)$ giving$7=-\frac{1}{2}\times 2+c$. Hence, $c=8$ and the equation of the line is $y=-\frac{1}{2}x+8$. Multiplying through by two to make all coefficients integer gives $2y=-x+16$. Finally, we have the equation in the required form: $x+2y-8=0$.

Always draw sketches to accompany your working – they help with sense checking.

### (Perpendicular)

The line $l_1$ has equation $y-2=\frac{1}{3}(x+2)$ relative to the origin $O$. The line $l_2$ is perpendicular to $l_1$, intersecting it at the point $A$, and also crosses the $x$-axis at the point $B(5,0)$. Find the distance $OA$ and area of the triangle $OAB$.

We start by sketching the information as this will tell us more about the problem. To find the distance $OA$ we require the coordinates of point $A$. To do this, we must first find the equation of line $l_2$. The line $l_1$ can be rewritten as $y=\frac{1}{3}x+\frac{8}{3}$, which has gradient 1/3. The gradient of $l_2$ is thus -3 and if it intersects the $x$-axis at 5, it must intersect the $y$-axis at 15, using rise over run. Hence, $l_2$ has equation $y=-3x+15$. To find the coordinates of $A$, the intersection of the two lines, we solve simultaneous equations: $y=\frac{1}{3}x+\frac{8}{3}$ and [math]y=-3x+15[/math], to get $x=3.7$ and $y=3.9$. See Simultaneous Equations.

The length $OA$ can now be found using Pythagoras: $OA=\sqrt{3.7^2+3.9^2}=5.38$ to two decimal places. The area of triangle $OAB$ can now be found using the height of the triangle (3.9) and its base (5), i.e. $0.5\times 5\times 3.9=9.75$.

#### (Proportionality)

Recall that the circumference of a circle is found from multiplying the diameter of the circle by $\pi$. Zero diameter means zero circumference and so the plot of circumference against diameter goes through the origin. The circumference of a circle is thus proportional to the diameter and the constant of proportion is $\pi$. Note that circumference is also proportional to radius but the constant of proportionality is then $2\pi$.

Note that this isn’t quite as simple as $y=2x$, for example, but $\pi$ is still a constant all the same.

#### (Proportionality)

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

$^\circ C$ | 0 | 10 | 20 | 30 | 40 |

$^\circ F$ | 32 | 50 | 68 | 86 | 104 |

From the graph we can see that the relationship between Celsius and Fahrenheit is precisely a linear. More specifically, the gradient is 9/5 and the y-intercept is 32 and so $F=\frac{9}{5}C+32$:

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