Sketch and analyse linear functions of the form $y = mx + c$, including finding gradient, $x$- and $y$-intercepts, equations of parallel and perpendicular lines, and solving linear equations and inequalities

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 $y = mx + c$:

• IMATH_4 = gradient (slope) = rise/run.
• IMATH_5 = $y$-intercept (the value of $y$ when $x = 0$).

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

IMATH_11 -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 $x \ge -3$.

Worked example

Find the equation of the line through $(-1, 4)$ and $(3, -2)$.

$m = (-2 - 4)/(3 - (-1)) = -6/4 = -3/2$.

Point-slope: $y - 4 = -\frac{3}{2}(x - (-1)) = -\frac{3}{2}(x + 1)$.

$y = -\frac{3}{2}x - \frac{3}{2} + 4 = -\frac{3}{2}x + \frac{5}{2}$.

Common traps

Using $x_1 - x_2$ in the denominator with $y_2 - y_1$ in the numerator. The differences must be in the same order.

Forgetting the sign in $y - y_1 = m(x - x_1)$. $y_1 = -3$ gives $y - (-3) = y + 3$.

Reversing the inequality only sometimes. Reverse only when multiplying or dividing by a negative.

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 $-1$.