Straight-line graphs for IGCSE Mathematics - Additional: Methods and Worked Examples

Where straight-line graphs sit within Additional Mathematics

Coordinate geometry occupies a foundational position in the IGCSE Additional Mathematics syllabus. While the topic may appear elementary at first glance, particularly for students who have already encountered y = mx + c in standard Mathematics (0580), the 0606 treatment extends considerably beyond basic gradient-intercept work. The distinguishing feature is the linear law: a technique that transforms non-linear relationships into straight-line form, enabling the determination of unknown constants from experimental or theoretical data. This single skill connects algebraic manipulation, logarithmic functions, and graphical interpretation in a way that few other topics achieve.

The syllabus objectives divide neatly into two tiers. The first tier consolidates and deepens line geometry: equations in multiple forms, parallelism, perpendicularity, midpoints, distances, and perpendicular bisectors. The second tier introduces the linear law, which appears with striking regularity on both Paper 1 and Paper 2.

Forms of the straight-line equation

Two standard forms appear throughout the IGCSE Additional Mathematics course, each suited to different problem contexts.

The gradient m of a line passing through two points (x₁, y₁) and (x₂, y₂) is calculated as m = (y₂ - y₁) / (x₂ - x₁). This ratio, sometimes called the rate of change, measures how steeply the line rises or falls per unit of horizontal displacement.

Worked example 1: Finding the equation from two points

Find the equation of the line through A(2, 5) and B(6, -3).

1. Calculate the gradient: m = (-3 - 5) / (6 - 2) = -8 / 4 = -2
2. Substitute into the point-slope form using point A: y - 5 = -2(x - 2)
3. Expand and rearrange: y - 5 = -2x + 4, so y = -2x + 9

Either point may be used in step 2; the final equation is the same regardless of which you choose. Using point B as a check: -2(6) + 9 = -12 + 9 = -3, which matches the y-coordinate of B.

Parallel and perpendicular lines

The conditions governing parallel and perpendicular lines are algebraically concise but conceptually important.

The perpendicular condition can be restated as m₂ = -1/m₁. This reciprocal-and-negate operation is worth committing to memory, as it appears in problems involving tangents, normals, and perpendicular bisectors alike.

Worked example 2: Finding a perpendicular line

Line L has equation y = 3x - 7. Find the equation of the line perpendicular to L that passes through the point (9, 2).

1. The gradient of L is 3. The perpendicular gradient is -1/3.
2. Using point-slope form: y - 2 = -1/3 (x - 9)
3. Expanding: y - 2 = -x/3 + 3, so y = -x/3 + 5

Midpoint, length, and perpendicular bisector

Three formulae underpin this section. For two points A(x₁, y₁) and B(x₂, y₂):

• Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
• Length AB = sqrt((x₂ - x₁)² + (y₂ - y₁)²)
• Perpendicular bisector: the line passing through M with gradient -1/m_AB

The perpendicular bisector is the locus of all points equidistant from A and B. Exam questions frequently ask candidates to find its equation, which requires combining the midpoint formula with the perpendicular gradient condition.

Worked example 3: Perpendicular bisector

Find the equation of the perpendicular bisector of the segment joining P(1, 4) and Q(7, -2).

1. Midpoint M = ((1+7)/2, (4+(-2))/2) = (4, 1)
2. Gradient of PQ = (-2 - 4)/(7 - 1) = -6/6 = -1
3. Perpendicular gradient = -1/(-1) = 1
4. Equation through M with gradient 1: y - 1 = 1(x - 4), giving y = x - 3

A useful verification step: confirm that P and Q are equidistant from any point on the bisector. Taking (0, -3) on the line y = x - 3, the distance to P is sqrt(1 + 49) = sqrt(50), and the distance to Q is sqrt(49 + 1) = sqrt(50). The distances match, confirming the bisector is correct.

The linear law: transforming curves into lines

This section represents the most distinctive and most frequently examined aspect of straight-line graphs within 0606. The principle is elegant: if a relationship between two variables is non-linear, a suitable substitution or logarithmic transformation can convert it into a linear form Y = mX + c, where Y and X are transformed variables. Plotting Y against X then yields a straight line whose gradient and intercept reveal the unknown constants.

The two standard models

The distinction between these two models is critical. In the power model y = Axⁿ, both variables are logged. In the exponential model y = Ab^x, only y is logged. Confusing the two is one of the most common errors candidates make, and it leads to an entirely incorrect identification of the axes.

Worked example 4: Power model

The variables x and y are connected by y = Axⁿ. When log y is plotted against log x, the resulting straight line passes through (0.5, 2.1) and (2.0, 5.6). Find the values of A and n.

1. The linear form is log y = n log x + log A. The gradient n = (5.6 - 2.1)/(2.0 - 0.5) = 3.5/1.5 = 7/3
2. Using the point (0.5, 2.1): 2.1 = (7/3)(0.5) + log A
3. 2.1 = 7/6 + log A, so log A = 2.1 - 7/6 = 2.1 - 1.1667 = 0.9333
4. A = 10^0.9333 = 8.58 (to 3 significant figures)

Worked example 5: Exponential model

Variables x and y satisfy y = Ab^x. The graph of ln y against x is a straight line with gradient 0.4 and passing through the point (0, 1.6). Find A and b.

1. The linear form using natural log is ln y = x ln b + ln A
2. The gradient is ln b = 0.4, so b = e^0.4 = 1.492 (to 4 significant figures)
3. At x = 0: ln y = ln A = 1.6, so A = e^1.6 = 4.953

Note that the choice of logarithm base (common log versus natural log) affects the numerical values of the gradient and intercept but not the method. Exam questions specify which base to use, either through the axis labels (log y vs ln y) or explicitly in the question text.

Beyond the two standard models

Some exam questions present relationships that require algebraic rearrangement before any logarithmic transformation. Consider y = a/x + bx. Multiplying both sides by x gives xy = a + bx². This is now in the form Y = mX + c with Y = xy, X = x², gradient b, and intercept a. Plotting xy against x² produces a straight line from which both constants can be read.

The general strategy for these non-standard forms is:

1. Identify the two unknown constants to be determined
2. Rearrange the equation so that one side contains a single expression in y (possibly combined with x) and the other side is linear in some function of x
3. Define Y and X as the transformed variables
4. Read the gradient and intercept from the transformed straight-line graph

Common errors and how to avoid them

Self-check questions

Test your understanding with these practice problems. Work through each one fully before checking.

1. Find the equation of the line passing through (-3, 8) and (5, -4). Express your answer in the form y = mx + c.
2. The line L₁ has equation 2y + 5x = 10. Find the gradient of L₁, then find the equation of the line L₂ that is perpendicular to L₁ and passes through (4, -1).
3. A and B have coordinates (2, 7) and (10, 3) respectively. Find the equation of the perpendicular bisector of AB.
4. Variables x and y are related by y = px^q. The graph of lg y against lg x is a straight line passing through (1, 3.5) and (3, 7.5). Find the values of p and q.
5. The relationship between x and y is y = Ke^rx. When ln y is plotted against x, the resulting line has gradient 2.5 and passes through (0, 3.2). Find K and r.

Exam strategy for straight-line graph questions

Questions on this topic in the IGCSE Additional Mathematics papers tend to fall into two categories. The first category tests line geometry directly: given points or equations, find gradients, equations of parallel or perpendicular lines, midpoints, distances, or perpendicular bisectors. These questions reward accuracy and careful sign management. The second category tests the linear law, and these questions typically carry higher marks because they combine algebraic transformation with graphical interpretation.

For linear law questions, a disciplined approach yields consistent results. Write out the full logarithmic transformation before attempting to read from the graph. Label clearly which transformed variable sits on which axis. Extract the gradient using two well-separated points from the line (not data points that may lie slightly off the line). Calculate the intercept algebraically from the gradient and one point, rather than attempting to read it from the graph, as the intercept may lie outside the plotted region.

Candidates who master the linear law gain access to marks that many of their peers leave on the table. The technique itself is mechanical once understood, but it requires precision at every step: the correct transformation, the correct axis assignment, accurate gradient calculation, and the final conversion from log form back to the original constants. Each step is a potential source of error, and each error propagates through the remainder of the solution. Methodical, clearly laid-out working is the most reliable safeguard.