National 5 Maths
Line of Best Fit

We can see from the graph that the points (2,3) and (9,7) are on the line of best fit.

Using our knowledge from the straight line topic:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{7-3}{9-2}\\[6pt] &=&\ \frac{4}{7} \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (2,3) to find the equation:
$$ \begin{eqnarray} y-b &=& m(x-a) \\[6pt] y-3 &=& \small{\frac{4}{7}}\normalsize (x-2) \\[6pt] 7(y-3) &=& 4(x-2) \\[6pt] 7y-21 &=& 4x-8 \\[6pt] -4x+7y &=& -8+21 \\[6pt] -4x+7y &=& 13 \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  We can see from the graph that the points (3,9) and (5,4) are on the line of best fit.

$$ \begin{eqnarray} m\ &=&\ \frac{Q_2-Q_1}{P_2-P_1}\\[6pt] &=&\ \frac{4-9}{5-3}\\[6pt] &=&\ -\frac{5}{2} \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (3,9) to find the equation:
$$ \begin{eqnarray} Q-b &=& m(P-a) \\[6pt] Q-9 &=& -\small{\frac{5}{2}}\normalsize (P-3) \\[6pt] 2(Q-9) &=& -5(P-3) \\[6pt] 2Q-18 &=& -5P+15 \\[6pt] 5P+2Q &=& 15+18 \\[6pt] 5P+2Q &=& 33 \\[6pt] \end{eqnarray} $$

(b)  We substitute \(P\) = 4 into the equation:

$$ \begin{eqnarray} 5P+2Q &=& 33 \\[6pt] 5(4)+2Q &=& 33 \\[6pt] 20+2Q &=& 33 \\[6pt] 2Q &=& 33-20 \\[6pt] 2Q &=& 13 \\[6pt] Q &=& \frac{13}{2} \\[6pt] \end{eqnarray} $$

(b)  We substitute \(Q\) = 3 into the equation:

$$ \begin{eqnarray} 5P+2Q &=& 33 \\[6pt] 5P+2(3) &=& 33 \\[6pt] 5P+6 &=& 33 \\[6pt] 5P &=& 33-6 \\[6pt] 5P &=& 27 \\[6pt] P &=& \frac{27}{5} \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  \(D\) is (3,100) and \(E\) is (15,340) so:

$$ \begin{eqnarray} m\ &=&\ \frac{W_2-W_1}{A_2-A_1}\\[6pt] &=&\ \frac{340-100}{15-3}\\[6pt] &=&\ \frac{240}{12}\\[6pt] &=&\ 20 \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (3,100) to find the equation:
$$ \begin{eqnarray} W-b &=& m(A-a) \\[6pt] W-100 &=& 20(A-3) \\[6pt] W-100 &=& 20A-60 \\[6pt] W &=& 20A-60+100 \\[6pt] W &=& 20A+40 \end{eqnarray} $$

(b)  One year is 12 months, so we substitute \(A\) = 12 into our equation:

$$ \begin{eqnarray} W &=& 20A+40 \\[6pt] &=& 20(12)+40 \\[6pt] &=& 240+40 \\[6pt] &=& 280 \\[6pt] \end{eqnarray} $$

So the one year old calf weighs 280 kg.

Video solution by Clelland Maths 

(a)  \(A\) is (8,14) and \(B\) is (12,20) so:

$$ \begin{eqnarray} m\ &=&\ \frac{P_2-P_1}{d_2-d_1}\\[6pt] &=&\ \frac{20-14}{12-8}\\[6pt] &=&\ \frac{6}{4}\\[6pt] &=&\ \frac{3}{2}\\[6pt] \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (8,14) to find the equation:
$$ \begin{eqnarray} P-b &=& m(d-a) \\[6pt] P-14 &=& \small\frac{3}{2}\normalsize(d-8) \\[6pt] 2(P-14) &=& 3(d-8) \\[6pt] 2P-28 &=& 3d-24 \\[6pt] 2P &=& 3d-24+28 \\[6pt] 2P &=& 3d+4 \\[6pt] \end{eqnarray} $$

(b)  We substitute \(d\) = 5 into our equation:

$$ \begin{eqnarray} 2P &=& 3d+4 \\[6pt] 2P &=& 3(5)+4 \\[6pt] 2P &=& 15+4 \\[6pt] 2P &=& 19 \\[6pt] P &=& \frac{19}{2} \\[6pt] P &=& 9.5 \\[6pt] \end{eqnarray} $$

So price of a 5 mile journey is £9.50.

Video solution by Clelland Maths 

(a)  From the graph, we can see two points that appear to be on the lines: (1.5,14) and (3.5,8). So we can find the gradient of the line of best fit.

$$ \begin{eqnarray} m\ &=&\ \frac{F_2-F_1}{E_2-E_1}\\[6pt] &=&\ \frac{14-8}{1.5-3.5}\\[6pt] &=&\ \frac{6}{-2}\\[6pt] &=&\ -\!3\\[6pt] \end{eqnarray} $$

Now we can use (\(a\),\(b\)) = (1.5,14) to find the equation:
$$ \begin{eqnarray} F-b &=& m(E-a) \\[6pt] F-14 &=& -\!3\normalsize(E-1.5) \\[6pt] F-14 &=& -\!3E+4.5 \\[6pt] F &=& -\!3E+4.5+14 \\[6pt] F&=& -\!3E+18.5 \\[6pt] \end{eqnarray} $$

(b)  We substitute \(E\) = 1.1 into our equation:

$$ \begin{eqnarray} F &=& -\!3E+18.5 \\[6pt] &=& -\!3(1.1)+18.5 \\[6pt] &=& -\!3.3+18.5 \\[6pt] &=& 15.2 \\[6pt] \end{eqnarray} $$

So Amaar should expect his car to get 15.2 kilometres per litre of fuel.

Video solution by Clelland Maths 

(a)  From the graph, we can read off two points: (5,20 000) and (25,50 000). So we can find the gradient of the line of best fit.

$$ \begin{eqnarray} m\ &=&\ \frac{P_2-P_1}{T_2-T_1}\\[6pt] &=&\ \frac{50\,000-20\,000}{25-5}\\[6pt] &=&\ \frac{30\,000}{20}\\[6pt] &=&\ 1500\\[6pt] \end{eqnarray} $$

Now we can use (\(a\),\(b\)) = (5,20 000) to find the equation:
$$ \begin{eqnarray} P-b &=& m(T-a) \\[6pt] P-20\,000 &=& 1500(T-5) \\[6pt] P-20\,000 &=& 1500T-7500 \\[6pt] P &=& 1500T-7500+20\,000 \\[6pt] P &=& 1500T+12\,500 \\[6pt] \end{eqnarray} $$

(b)  We substitute \(T\) = 8 into our equation:

$$ \begin{eqnarray} P &=& 1500T+12\,500 \\[6pt] &=& 1500(8)+12\,500 \\[6pt] &=& 12\,000+12\,500 \\[6pt] &=& 24\,500 \\[6pt] \end{eqnarray} $$

So an employee who has worked for the business for 8 years is likely to earn around £24 500.

Video solution by Clelland Maths