Year 9 Maths Advanced Coordinate Geometry and Graphs

The Gradient Intercept Equation

Q: What does y = mx + c mean?

It is the equation of a straight line. m is the gradient, how much y changes when x goes up by 1, and c is the y-intercept, where the line crosses the y-axis.

Q: How do you find the gradient and y-intercept from an equation?

Rearrange the equation to y = mx + c. The coefficient of x is the gradient and the constant term is the y-intercept.

Q: How do you graph a line using the gradient and y-intercept?

Plot the y-intercept on the y-axis. Write the gradient as a fraction, rise over run, and step from the intercept that far across and up or down. Rule a line through the points.

Q: How do you tell if two lines are perpendicular?

Multiply their gradients. If the product is negative 1, the lines are perpendicular. For example, gradients 3 and negative one third are perpendicular.

20 practice questions 1 video lesson Theory + worked examples

When a line is written as $y = mx + c$, called gradient-intercept form, $m$ is the gradient and $c$ is the $y$-intercept, so a line's equation can be read straight from its graph. This NSW Year 9 Advanced maths subtopic uses this form (NESA outcome MA5‑LIN‑C‑02).

You will state the gradient and $y$-intercept, match an equation to its graph, and write the equation of a graphed line. You will rearrange equations such as $2x + 3y = 6$ and $px + qy + r = 0$ into this form, write a parallel line, show that two lines are perpendicular, model a burning candle, and find the right-angled triangle formed by three lines. It is part of Year 9 Advanced Coordinate Geometry and Graphs.

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Practice questions

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Theory

In the gradient–intercept form $y=mx+c$, the number multiplying $x$ is the gradient $m$ and the constant $c$ is the $y$-intercept. This Year 9 Advanced maths page shows how to rearrange any linear equation into this form, graph it from $m$ and $c$, test for parallel and perpendicular lines, and explain $m$ and $c$ in context.

Any non-vertical straight line can be written as $y=mx+c$. This is the gradient–intercept form.

$m$ is the gradient: the change in $y$ for each increase of $1$ in $x$. $c$ is the $y$-intercept: the value of $y$ where the line crosses the $y$-axis, at $(0,c)$.

The equation must start $y=$ before you read anything. In $y=5-2x$ the gradient is $-2$, the number multiplying $x$, and $c=5$.

Lines with the same gradient are parallel. Two lines are perpendicular when the product of their gradients is $-1$: each gradient is the negative reciprocal of the other.

In real situations, $c$ is the starting value (when the input is $0$) and $m$ is the rate of change, with units.

In $y = mx + c$, the line crosses the $y$-axis at $c$, and $m$ is the rise for each run of $1$.

Same gradient, different $c$: parallel lines. Gradients $\dfrac{1}{2}$ and $-2$ multiply to $-1$, so the red line is perpendicular to the navy lines.

\[y = mx + c\]

y = m x + c

Relationship	Gradients
parallel	$m_1 = m_2$
perpendicular	$m_1 \times m_2 = -1$, so $m_2 = -\dfrac{1}{m_1}$

From the general form $ax+by+c=0$ (with $b\ne 0$):

\[y = -\dfrac{a}{b}x - \dfrac{c}{b}\]

y = - \frac{a}{b} x - \frac{c}{b}

Clear fractions first. For an equation such as \(\dfrac{x}{4}-\dfrac{y}{6}=1\), multiply every term by \(12\) to get \(3x-2y=12\), then make \(y\) the subject.

Point–gradient form. \(y-y_1=m(x-x_1)\). Expand the bracket and collect constants to reach \(y=mx+c\).

How to read and graph y = mx + c

Rearrange so the equation starts $y=$: move the $x$-term and constant across, clear any fractions, then divide every term by the coefficient of $y$.
Read $m$ (the coefficient of $x$, with its sign) and $c$ (the constant).
Plot $(0,c)$. From there use the gradient as a fraction: run across, rise up (or down if negative).
Rule the line through the two points and check one more point by substitution.

Problems built on y = mx + c

Write the equation from a given gradient and $y$-intercept, or read both from a graph: $c$ where it crosses the $y$-axis, $m$ from rise over run between two grid points.
Which is steeper? Compare the sizes of the gradients.
Parallel line: copy $m$, change $c$.
Perpendicular test: multiply the gradients; $-1$ means perpendicular (Example 2).
Unknown in a coefficient: find the gradient in terms of the unknown, then set it equal to the required gradient (Example 4).
Conditions on m and c: if a line passes through a point, substitute it into $y=mx+c$; a second fact, such as $c=3m$, gives a second equation.
Context: rearrange to the form output $=m\times$ input $+c$; $c$ is the starting value and $m$ the rate (Example 3).

Example 1 — Rearrange, read, graph

Write $4x-3y=9$ in gradient–intercept form. State the gradient and $y$-intercept, and graph the line.

Solution

Make $y$ the subject. Move the $x$-term first, then divide every term by the coefficient of $y$.

$-3y$	$=$	$-4x+9$
$y$	$=$	$\dfrac{-4x}{-3}+\dfrac{9}{-3}$
$y$	$=$	$\dfrac{4}{3}x-3$

y = \frac{4}{3} x - 3

So $m=\mathbf{\dfrac{4}{3}}$ and $c=\mathbf{-3}$. Plot $(0,-3)$, then go $3$ across and $4$ up to $(3,1)$.

Dividing by $-3$ changes the sign of every term. That is why the gradient is positive.

Example 2 — Parallel or perpendicular?

Compare the lines $L_1:\ y=3x-2$, $L_2:\ 6x-2y=-5$ and $L_3:\ x+3y=9$. Which pairs are parallel and which are perpendicular?

Solution

Put each line in the form $y=mx+c$ and compare gradients.

$L_2:\ \ -2y$	$=$	$-6x-5$
$y$	$=$	$3x+\dfrac{5}{2}$
$L_3:\ \ 3y$	$=$	$-x+9$
$y$	$=$	$-\dfrac{1}{3}x+3$

$m_1 = 3$, $m_2 = 3$, $m_3 = -\dfrac{1}{3}$.

$m_1 \times m_3$

$=$

$3\times\left(-\dfrac{1}{3}\right) = -1$

m_{1} m_{3} = - 1

$L_1$ and $L_2$ are parallel (equal gradients, different intercepts). $L_3$ is perpendicular to both.

Example 3 — Reading m and c in a context

The cost, $\$C$, of hiring a hall for $h$ hours satisfies $2C-90h=300$.
(i) Write the equation in the form $C=mh+c$.
(ii) What do $m$ and $c$ mean?
(iii) Find the cost of a $5$-hour hire.

Solution

(i) Make $C$ the subject.

$2C$	$=$	$90h+300$
$C$	$=$	$45h+150$

C = 45 h + 150

(ii) $c = 150$: a fixed booking fee of $\$150$, paid even for $0$ hours. $m = 45$: each extra hour adds $\$45$. The gradient is a rate, in dollars per hour.

(iii)

$C$

$=$

$45\times 5+150 = \mathbf{375}$

A $5$-hour hire costs $\$375$.

Example 4 — A perpendicular condition

The line $(2k+1)x-4y=8$ is perpendicular to $y=-2x+3$. Find the value of $k$.

Solution

Rearrange to find the gradient in terms of $k$.

$-4y$	$=$	$-(2k+1)x+8$
$y$	$=$	$\dfrac{2k+1}{4}x-2$

Perpendicular gradients multiply to $-1$, so this gradient must be $\dfrac{1}{2}$, the negative reciprocal of $-2$.

$\dfrac{2k+1}{4}$	$=$	$\dfrac{1}{2}$
$2k+1$	$=$	$2$
$k$	$=$	$\mathbf{\dfrac{1}{2}}$

\frac{2 k + 1}{4} = \frac{1}{2}

Check: with $k=\dfrac{1}{2}$ the line is $2x-4y=8$, or $y=\dfrac{1}{2}x-2$, and $\dfrac{1}{2}\times(-2) = -1$.

Common pitfalls

Get y on its own first. In $4x-3y=9$ the gradient is not $4$. Rearranging gives $y=\dfrac{4}{3}x-3$, so $m=\dfrac{4}{3}$.

Divide every term. From $-3y=-4x+9$, dividing only the $x$-term by $-3$ leaves the wrong intercept. Each term must be divided.

The sign belongs to the number. In $y=5-2x$, $m=-2$, not $2$, and $c=5$. Reorder to $y=-2x+5$ if it helps.

Negative reciprocal, both parts. Perpendicular to $m=-2$ is $\dfrac{1}{2}$: flip the fraction and change the sign. Doing only one of the two gives $-\dfrac{1}{2}$ or $2$.

Frequently asked questions

What does y = mx + c mean?

It is the equation of a straight line. $m$ is the gradient, how much $y$ changes when $x$ goes up by $1$, and $c$ is the $y$-intercept, where the line crosses the $y$-axis.

How do you find the gradient and y-intercept from an equation?

Rearrange the equation to $y=mx+c$. The coefficient of $x$ is the gradient and the constant term is the $y$-intercept.

How do you graph a line using the gradient and y-intercept?

Plot the $y$-intercept on the $y$-axis. Write the gradient as a fraction, rise over run, and step from the intercept that far across and up or down. Rule a line through the points.

How do you tell if two lines are perpendicular?

Multiply their gradients. If the product is $-1$, the lines are perpendicular. For example, gradients $3$ and $-\dfrac{1}{3}$ are perpendicular.

Can a vertical line be written as y = mx + c?

No. A vertical line has an undefined gradient, so it is written $x=k$ instead.

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Graphing Linear Equations

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Finding the Equation of a Line

\(-3y\)	\(=\)	\(-4x+9\)
\(y\)	\(=\)	\(\dfrac{-4x}{-3}+\dfrac{9}{-3}\)
\(y\)	\(=\)	\(\dfrac{4}{3}x-3\)

\(L_2:\ \ -2y\)	\(=\)	\(-6x-5\)
\(y\)	\(=\)	\(3x+\dfrac{5}{2}\)
\(L_3:\ \ 3y\)	\(=\)	\(-x+9\)
\(y\)	\(=\)	\(-\dfrac{1}{3}x+3\)

\(2C\)	\(=\)	\(90h+300\)
\(C\)	\(=\)	\(45h+150\)

\(-4y\)	\(=\)	\(-(2k+1)x+8\)
\(y\)	\(=\)	\(\dfrac{2k+1}{4}x-2\)

\(\dfrac{2k+1}{4}\)	\(=\)	\(\dfrac{1}{2}\)
\(2k+1\)	\(=\)	\(2\)
\(k\)	\(=\)	\(\mathbf{\dfrac{1}{2}}\)

Relationship	Gradients
parallel	\(m_1 = m_2\)
perpendicular	\(m_1 \times m_2 = -1\), so \(m_2 = -\dfrac{1}{m_1}\)

The Gradient Intercept Equation

📖 Theory

How to read and graph y = mx + c

Problems built on y = mx + c

Common pitfalls

Frequently asked questions

Theory