Line of Best Fit

Video lesson · from 11:274 worked examples

One lesson video covers all of Statistics, so it opens at 11:27 for this topic — not from the beginning.

Theory

A scattergraph plots data points to show the relationship between two variables. You must be able to draw a straight "line of best fit" through the data.

Determining the Equation

To find the equation of your line of best fit, you select two points that lie exactly on the line (not necessarily data points). You then calculate the gradient ( $m$ ) and use $y - b = m(x - a)$ to find the equation in terms of the variables on the axes.

Estimating

Once you have the linear model (the equation), you must be able to use it to estimate a $y$ -value for a given $x$ -value (or use transposition to find $x$ given $y$ ).

The Golden Rule: pick two points that lie exactly on the line (not raw data points), find the gradient, then use $y - b = m(x - a)$ — and rename $x$ and $y$ to the letters on the axes. To estimate, substitute into the finished equation; to reverse it, substitute and rearrange.

⚠️ Common Examiner Traps

Points on the line: read the two points off the drawn line where it crosses grid corners — do not use scattered data points.
Rename the variables: the answer must be in the question's letters (e.g. $W$ and $A$ ), not $x$ and $y$ .
Negative gradient: a downward line has a negative gradient — keep the sign through the working.
Simplest form: tidy the equation fully; often the constant cancels to give a clean result.

Worked examples

Example 1

A scattergraph shows the relationship between temperature ( $T$ in °C) and ice cream sales ( $S$ in £). A line of best fit is drawn, passing exactly through the points (10, 150) and (20, 300). Find the equation of the line of best fit in terms of $S$ and $T$ .

Step 1: Find the gradient ( $m$ ):

m = \frac{300 - 150}{20 - 10} = \frac{150}{10} = 15

Step 2: Substitute $m = 15$ and the point (10, 150) into $y - b = m(x - a)$ . Remember to use $S$ instead of $y$ and $T$ instead of $x$ :

S - 150 = 15(T - 10)

Step 3: Expand and simplify:

S - 150 = 15T - 150 \Rightarrow S = 15T

Answer: $S = 15T$ .

Example 2

Using the equation derived above ( $S = 15T$ ), estimate the ice cream sales on a day when the temperature is 25°C.

Step 1: Substitute $T = 25$ into the equation:

S = 15(25)

Step 2: Calculate the result:

S = 375

Answer: Estimated sales are £375.

Example 3

A scattergraph shows the relationship between the age of a car ( $A$ in years) and its value ( $V$ in £). A line of best fit passes exactly through the points (2, 8000) and (5, 3500). Find the equation of the line of best fit in terms of $V$ and $A$ .

Step 1: Find the gradient ( $m$ ):

m = \frac{3500 - 8000}{5 - 2} = \frac{-4500}{3} = -1500

Step 2: Substitute $m = -1500$ and the point (2, 8000) into the straight line equation. Use $V$ for $y$ and $A$ for $x$ :

V - 8000 = -1500(A - 2)

Step 3: Expand and simplify:

V - 8000 = -1500A + 3000

V = -1500A + 11000

Answer: $V = -1500A + 11000$ .

Example 4

🔗 Bringing it together (reverse estimate)

Using the equation $V = -1500A + 11000$ from above, estimate the age of a car whose value is £5000.

Step 1: This time the value $V$ is known, so substitute $V = 5000$ and solve for $A$ :

5000 = -1500A + 11000

Step 2: Rearrange — subtract 11000 from both sides:

-6000 = -1500A

Step 3: Divide by $-1500$ :

A = 4

Answer: the car is estimated to be 4 years old.

Ready to practise?

7 Line of Best Fit questions with answers and video solutions