Linear functions and graphs (VCE Maths Methods Unit 1)

What this dot point is asking

VCAA wants you to sketch and analyse linear functions $y = mx + c$, find gradient and intercepts, write equations from given conditions, and use parallel and perpendicular gradient relationships.

Gradient and intercept

For $y = mx + c$:

• $m$ = gradient (slope) = rise/run.
• $c$ = $y$-intercept (the value of $y$ when $x = 0$).

Gradient from two points $(x_1, y_1)$ and $(x_2, y_2)$:

$x$-intercept

The $x$-intercept is where the line crosses the $x$-axis ($y = 0$).

$0 = mx + c$, so $x = -c/m$.

Point-slope form

A line of gradient $m$ passing through $(x_1, y_1)$:

Parallel and perpendicular lines

Two lines with gradients $m_1$ and $m_2$:

• Parallel: $m_1 = m_2$.
• Perpendicular: $m_1 m_2 = -1$, equivalently $m_2 = -1/m_1$.

Horizontal lines ($m = 0$) and vertical lines ($m$ undefined) are perpendicular to each other.

Solving linear equations and inequalities

Solve $3x - 5 = 7$: add $5$, divide by $3$: $x = 4$.

For inequalities, the inequality direction reverses when multiplying or dividing by a negative number. $-2x \le 6$ becomes $x \ge -3$.

Midpoint, distance and the perpendicular bisector

Linear coordinate geometry in Unit 1 also draws on two formulas for a segment joining $(x_1, y_1)$ and $(x_2, y_2)$. The midpoint is $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$, and the distance is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. The perpendicular bisector of a segment passes through its midpoint with gradient equal to the negative reciprocal of the segment's gradient; it is the set of points equidistant from the two endpoints, and is a recurring exam construction.

Linear simultaneous equations

Two lines either intersect once (different gradients), never (parallel, equal gradients but different intercepts), or coincide (identical lines). To find an intersection, solve the pair simultaneously by substitution or elimination. For example, $y = 2x + 1$ and $y = -x + 7$ give $2x + 1 = -x + 7$, so $3x = 6$, $x = 2$, and $y = 5$: the lines meet at $(2, 5)$. Recognising the no-solution and infinite-solution cases from the gradients is a common short-answer item.

In one sentence

Linear functions $y = mx + c$ have gradient $m$ (rise over run, found from two points by $(y_2 - y_1)/(x_2 - x_1)$) and $y$-intercept $c$; parallel lines share a gradient and perpendicular lines have gradients whose product is $-1$.

Examples in context

Example 1. A phone plan as a linear model. A mobile plan charges a $\$20$ monthly fee plus $\$0.15$ per gigabyte of data, so the monthly cost is $C = 0.15x + 20$ where $x$ is gigabytes used. The gradient $0.15$ is the cost per extra gigabyte and the $y$-intercept $20$ is the base fee. A month using $30$ GB costs $C = 0.15(30) + 20 = \$24.50$. If a bill is $\$26$, then $0.15x + 20 = 26$ gives $x = \frac{6}{0.15} = 40$ GB.

Example 2. Perpendicular path of a delivery drone. A drone flies along $y = 2x + 1$. To return perpendicularly from the point $(3, 7)$, the new path has gradient $-\frac{1}{2}$ (negative reciprocal). Point-slope form: $y - 7 = -\frac{1}{2}(x - 3)$, so $y = -\frac{1}{2}x + \frac{17}{2}$. The two paths meet at right angles since $2 \times (-\frac{1}{2}) = -1$.

Try this

Q1. Find the gradient and $x$-intercept of $y = 4x - 12$. [2 marks]

• Cue. Gradient $4$; $x$-intercept where $0 = 4x - 12$, so $x = 3$.

Q2. A taxi fare is $\$4.50$ flagfall plus $\$2.20$ per km. (a) Write the fare $F$ as a function of distance $d$. (b) Find the distance for a $\$26.50$ fare. [2+2 marks]

• Cue. (a) $F = 2.2d + 4.5$. (b) $2.2d + 4.5 = 26.5 \Rightarrow d = \frac{22}{2.2} = 10$ km.

Q3. Find the equation of the line through $(1, 4)$ parallel to $y = -3x + 5$. [2 marks]

• Cue. Same gradient $-3$: $y - 4 = -3(x - 1)$, so $y = -3x + 7$.