VCE Mathematical Methods (Year 11) · Unit 1 — Algebra & Structure · 12m read

Linear Relations and Equations

Ever wonder why graphs of straight lines show up everywhere from phone bills to fitness tracking? Linear equations are the foundation for understanding relationships between two variables, and mastering them means you can predict real-world outcomes using simple mathematical rules. You'll learn to solve for unknown values and create models that turn messy situations into neat, predictable patterns — a skill that's essential for VCE math and beyond.

Learning objective: Solve linear equations and model with linear functions.

By the end of this lesson you can:

• I solve linear equations algebraically.
• I find gradient and intercepts of a line.
• I use linear models to answer context questions.

A linear relation is a straight train track. Constant slope, no curves. Know the gradient and one point and you know the whole track.

Lines, gradients and equations

FORMS OF A LINE:
• GRADIENT-INTERCEPT: y = mx + c (m = gradient, c = y-intercept).
• POINT-GRADIENT: y − y₁ = m(x − x₁).
• GENERAL: ax + by + c = 0.

GRADIENT:
• m = (y₂ − y₁) / (x₂ − x₁) = rise/run.
• Positive → rising, negative → falling.
• Horizontal line: m = 0. Vertical line: m is undefined.

PARALLEL & PERPENDICULAR:
• Parallel lines share gradient: m₁ = m₂.
• Perpendicular lines: m₁ × m₂ = −1.

FINDING EQUATIONS:
• Given 2 points → compute m, substitute into y − y₁ = m(x − x₁).
• Given gradient + point → substitute into point-gradient form.
• Rearrange into required form.

SOLVING LINEAR EQUATIONS:
• Do the same thing to both sides.
• Collect like terms.
• Beware of fractions — multiply through by LCD.

LINEAR MODELLING:
• Identify independent (x) and dependent (y) variables.
• Gradient = RATE (e.g., $ per hour, km per litre).
• y-intercept = FIXED cost / starting value.

INEQUALITIES:
• Solve like equations BUT flip the inequality when multiplying/dividing by a negative.
• Graph on a number line (open circle < >, closed circle ≤ ≥).

DISTANCE & MIDPOINT (between (x₁,y₁) and (x₂,y₂)):
• Distance = √[(x₂−x₁)² + (y₂−y₁)²].
• Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).

Line essentials

• Gradient: rise ÷ run = m.
• Intercept: c = value of y when x = 0.
• Perpendicular: m₁ × m₂ = −1.

Linear relations

• Slope-intercept: y = mx + c
• Point-slope: y - y_1 = m(x - x_1)
• Distance: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Quick reference

| Form | Equation |
|---|---|
| Gradient-intercept | y = mx + c |
| Point-gradient | y − y₁ = m(x − x₁) |
| General | ax + by + c = 0 |
| Gradient | m = (y₂ − y₁)/(x₂ − x₁) |
| Parallel | Equal gradients |
| Perpendicular | m₁m₂ = −1 |

Worked example (Medium)

Find the equation of the line through (2, −1) perpendicular to y = 3x + 4.

1. Perpendicular gradient. m₁ = 3 → m₂ = −1/3.
2. Point-gradient. y − (−1) = −1/3 (x − 2) → y + 1 = −(x−2)/3.
3. Simplify. 3y + 3 = −x + 2 → x + 3y + 1 = 0.

Answer: y = −x/3 − 1/3 or x + 3y + 1 = 0.

Worked example (Hard)

The midpoint of segment AB is M(3, 1). If A = (−1, 4), find B and the equation of the perpendicular bisector of AB.

1. Find B using midpoint formula. (−1 + x₂)/2 = 3 → x₂ = 7. (4 + y₂)/2 = 1 → y₂ = −2. So B = (7, −2).
2. Gradient of AB. m = (−2 − 4)/(7 − (−1)) = −6/8 = −3/4.
3. Perpendicular bisector gradient. m_perp = 4/3 (negative reciprocal).
4. Equation through M(3, 1). y − 1 = 4/3(x − 3) → 3y − 3 = 4x − 12 → 4x − 3y − 9 = 0.

Answer: B = (7, −2). Perpendicular bisector: 4x − 3y − 9 = 0.

Common mistakes

• Forgetting to flip the inequality sign when multiplying or dividing by a negative Students apply the same rule as equations — keep the sign unchanged Fix: Always ask: "Am I multiplying/dividing by a negative?" If yes, reverse the inequality symbol.
• Subtracting coordinates in inconsistent order in the gradient formula Using (y₂ − y₁)/(x₁ − x₂) mixes the subscript order, giving the wrong sign Fix: Lock in top-to-bottom order: pick one point as "first" and subtract consistently.
• Leaving the answer in point-gradient form when general form is requested Students stop once they substitute into y − y₁ = m(x − x₁) Fix: Read the question for "general form" (ax + by + c = 0) and rearrange fully.

Why linear models dominate real-world applications

Linear relationships form the backbone of modelling in VCE Methods because they are the simplest functions that capture a constant rate of change. Whenever a quantity grows or shrinks by the same amount per unit of input, a straight line fits perfectly.

In economics, a linear cost function $C(x) = mx + c$ splits neatly into a variable cost (gradient $m$) and a fixed cost (intercept $c$). Revenue minus cost gives profit, and the break-even point is where revenue equals cost — a simultaneous-equation problem.

In science, many phenomena are approximately linear over small ranges. Hooke's law ($F = kx$) and Ohm's law ($V = IR$) are both linear models that work well within their valid domain, even though real springs and resistors deviate at extremes.

A key idea VCAA tests is that the gradient carries units — it is a rate. If the x-axis is time (hours) and the y-axis is distance (km), the gradient is speed (km/h). Interpreting the gradient and intercept in context is worth dedicated marks on almost every SAC.

Finally, two intersecting lines model constraints: supply vs demand, distance vs time for two travellers, budget limits. The algebra of simultaneous linear equations appears throughout Units 1–2 and beyond.

Practice questions

1. The gradient of the line 2x − 3y = 6 is:

   1. −2/3
   2. 2/3
   3. 3/2
   4. −3/2

   (1 marks)

2. A taxi charges a $4 flagfall plus $2.20 per km. Write a cost function C(d) for a trip of d km, and find the cost for an 8 km trip.

   (3 marks)

Summary

• Solve linear equations algebraically.
• Find gradient and intercepts of a line.
• Use linear models to answer context questions.