Modelling with straight lines

In a nutshell

A straight line graph can be used to model data, which allows you to visualise how two variables change in respect to each other. You can see if two variables are proportional to one another and can make future predictions.

Proportionality

Two variables are proportional when they have an unchanging relationship with each other. In straight line graphs, $y$ is proportional to $x$ . The gradient informs you how much $y$ changes in respect to $x$ . The way $y$ changes with respect to $x$ is constant for straight line graphs.

Maths; Straight line graphs; KS5 Year 12; Modelling with straight lines

If $x$ increases by $1$ unit, $y$ increases by $k$ units.

Proportionality allows you to make predictions. You can solve for a certain $y$ value given any $x$ value with the equation of a line.

Linear models

Linear modelling is a method to visualise data. If two variables are proportional to each other, you can plot them on a graph and draw a line connecting them. The line allows you to use linear equations, and therefore predict how the variables will continue to change.

procedure

1.	Plot the given data points on a graph.
2.	Draw a line of best fit that most closely fits the trend of the data.
3.	Use two points on the line to solve for the line's gradient and equation.

Note: Not all the points need to lie directly on the line for the method to still be valid. However, a linear model becomes less accurate the further away points are from the line.

Example 1

A cyclist is riding a bike when she passes by a tree. Her distance in metres is recorded in $1$ second intervals from that point onwards. Below is a table showing her distance from the tree every second.

Distance (metres)	$0$	$5$	$12$	$18$	$25$	$30$
Time (seconds)	$0$	$1$	$2$	$3$	$4$	$5$

Create a linear model of her speed. From that model, find out how many seconds will elapse before she makes it to the next tree,
$120$ metres away.

Plot out the data and draw a line of best fit.

Maths; Straight line graphs; KS5 Year 12; Modelling with straight lines

Choose two points on the line and find the gradient of the line.

$(2,12)\space(3,18)$

$x_1=2, y_1=12, x_2=3, y_2=18$

$\begin{aligned}\\m&=\dfrac{y_2-y_1}{x_2-x_1}\\m&=\dfrac{18-12}{3-2}\\m&=\dfrac{6}{1}\\m&=6\\\end{aligned}$

Use the gradient to find the equation of the line.

$\begin{aligned}y-y_1&=m(x-x_1)\\y-12&=6(x-2)\\y-12&=6x-12\\y&=6x\\\end{aligned}$

Solve for $x$  when $y=120$ .

$\begin{aligned}&y=120\\6&x=120\\&x=\dfrac{120}{6}\\&\underline{x=20\space seconds}\end{aligned}$