Additional Mathematics notes, chapter 6
Linear Law
Textbook Additional Mathematics Form 4 KSSM (DLP), pages 152–173
What this chapter covers
- Differentiate between linear and non-linear relations from tables of data and graphs (6.1.1)
- Draw lines of best fit for graphs of linear relations, using suitable scales (6.1.2)
- Form the equation of a line of best fit, Y = mX + c, from two points or from the graph (6.1.3)
- Interpret information from a line of best fit, by reading the graph or by using its equation (6.1.4)
- Convert non-linear equations to the linear form Y = mX + c and identify Y, X, m and c (6.2.1)
- Use linear law to find unknown constants of a non-linear relation from a straight line graph (6.2.1)
- Solve problems that involve linear law, including logarithmic forms (6.3.1)
6.1Linear and non-linear relations
Linear relation: the graph is a straight line, with equation Y = mX + c. Non-linear relation: the graph is a curve.
In Y = mX + c, X is the variable on the horizontal axis, Y is the variable on the vertical axis, m is the gradient and c is the Y-intercept.
A non-linear graph can become a straight line when you change the variable on the X-axis, the Y-axis, or both. For example, y = x² + 1 gives a curve for y against x but a straight line for y against x² (gradient 1, intercept 1).
Example 1 (textbook p. 154): linear or not?
Y = 3 − 2X gives the values X = −1, 0, 1, 2 → Y = 5, 3, 1, −1. Each time X goes up by 1, Y goes down by 2, so the points lie on a straight line: linear.Y = X² gives Y = 1, 0, 1, 4 for the same X. The change in Y is not constant, so the graph is a curve: non-linear.
The points (1, 5), (2, 7), (3, 9), (4, 11) are plotted. Is the relation linear? Give a reason.
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Linear. Y goes up by 2 for every 1 unit of X, so the points lie on a straight line (Y = 2X + 3).
Line of best fit
A line of best fit is a straight line that passes through, or as near as possible to, most of the plotted points. Points not on the line must be spread evenly on both sides of it.
- Choose scales so the points fill most of the graph paper. Use a uniform scale such as 1 cm to 1 unit or 2 cm to 5 units.
- Draw with a ruler; do not join the points one by one.
- A point that is far from the trend, for example one recorded wrongly, is marked and left out, as the line follows the trend of the rest (see Intensive Practice 6.3, Q2 on p. 168).
Equation of a line of best fit
Choose two points on the line (not data points that are off the line), then find m = (Y₂ − Y₁) ÷ (X₂ − X₁) and c from the Y-intercept, or by substituting one point into Y = mX + c.
Example 4 (textbook p. 158): equation of the line of best fit
Equation: y = 1.4x + 2.
Take the gradient from two points on the line, far apart, and read c where the line meets the vertical axis at X = 0. If the horizontal axis does not start at 0, find c by substitution instead.
Interpreting the graph
Use the line (or its equation) to find values that were not measured. Reading from the line within the plotted values is interpolating; extending the line beyond is extrapolating. The equation is more accurate than reading the graph.
Example 5 (textbook p. 159): reading a log graph
The line of best fit of log₁₀ y against x has gradient 2 and Y-intercept 0.2, so log₁₀ y = 2x + 0.2.When x = 0: log₁₀ y = 0.2, so y = 100.2 = 1.585.
When y = 20: log₁₀ 20 = 1.3, so 2x + 0.2 = 1.3 and x = 0.55.
A line of best fit has equation y = 1.4x + 2. Find y when x = 5 and x when y = 12.
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x = 5: y = 1.4(5) + 2 = 9.
y = 12: 1.4x = 10, so x = 7.14.
y = 12: 1.4x = 10, so x = 7.14.
6.2Linear law and non-linear relations
Linear law: a non-linear equation can be rewritten in the form Y = mX + c. Plot Y against X to get a straight line, then read m and c from the graph to find the unknown constants.
Rules for choosing X and Y: both must contain variables only, never the unknown constants. m and c must contain constants only. The same equation can be converted in more than one way.
Converting to the linear form
| Non-linear equation | Step | Y | X | m | c |
|---|---|---|---|---|---|
| y = ax² + b | already linear in x² | y | x² | a | b |
| y = ax + b/x | multiply by x: xy = ax² + b | xy | x² | a | b |
| y = ax + b/x | divide by x: y/x = b(1/x²) + a | y/x | 1/x² | b | a |
| y = px² + qx | divide by x: y/x = px + q | y/x | x | p | q |
| y = pqx | take log₁₀: log y = (log q)x + log p | log₁₀ y | x | log₁₀ q | log₁₀ p |
| y = axb | take log₁₀: log y = b log x + log a | log₁₀ y | log₁₀ x | b | log₁₀ a |
After taking logarithms, p and q are found from log₁₀ p and log₁₀ q by anti-log: p = 10c, q = 10m. Do not give log₁₀ q as the answer for q.
Example 7 (textbook p. 163): find p and q from a graph of y/x against x
3y − px² = qx. Divide by 3x: y/x = (p/3)x + q/3, so Y = y/x, X = x, m = p/3, c = q/3.From the line: Y-intercept = 22 and gradient = −2.
q/3 = 22, so q = 66. p/3 = −2, so p = −6.
Worked example: y = pqx
For y = 3 × 2x, log₁₀ y = (log₁₀ 2)x + log₁₀ 3 = 0.301x + 0.477.
Convert y = 5x² + 3x to the form Y = mX + c and identify Y, X, m and c. Then do the same for y = h xk.
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First: divide by x, y/x = 5x + 3. Y = y/x, X = x, m = 5, c = 3.
Second: log₁₀ y = k log₁₀ x + log₁₀ h. Y = log₁₀ y, X = log₁₀ x, m = k, c = log₁₀ h.
Second: log₁₀ y = k log₁₀ x + log₁₀ h. Y = log₁₀ y, X = log₁₀ x, m = k, c = log₁₀ h.
6.3Applications of linear law
Follow the textbook's four steps for a problem: understand (what are the variables and the axes), plan (convert to Y = mX + c, make a new table), implement (plot, read m and c, find the constants), conclude (check by substituting back).
Example 9 (textbook p. 166): P = 180 + rK − sK²
Rearrange: P − 180 = rK − sK². Divide by K: (P − 180)/K = −sK + r.So Y = (P − 180)/K, X = K, m = −s, c = r.
From the graph: c = 1.6, so r = 1.6; gradient −s = −0.56, so s = 0.56.
Check with K = 3: P = 180 + 1.6(3) − 0.56(9) = 179.76, matching the table (179.7).
When a formula has several terms with variables of different powers, move the term that does not contain constants to the left, then divide to leave a single constant term. Always make a new table of Y and X values before plotting.
The height h of an object is related to time t by h = at − bt². Which variables do you plot to get a straight line, and what do the gradient and intercept give?
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Divide by t: h/t = −bt + a. Plot h/t against t. The gradient is −b and the h/t-intercept is a.
Notes follow Additional Mathematics Form 4 KSSM (DLP), Chapter 6, pp. 152 to 173; figures are redrawn from the textbook examples.