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Year 11 Methods (Unit 1 & 2) Coordinate Geometry And Linear Relations

Graphing Straight Lines

20 practice questions 1 video lesson Theory + worked examples

Graphing straight lines is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). Because two points fix a line, a clear sketch needs only a couple of well-chosen points, such as where the line crosses each axis.

You will find the horizontal and vertical intercepts by setting the other variable to zero, sketch from the gradient and intercept, and link each line's algebraic and graphical representations with confidence.

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Theory

In Year 11 Mathematical Methods (QCAA), sketching a straight line means plotting a couple of key points and joining them. Two reliable methods are the gradient-intercept method (plot the \(y\)-intercept, then step by the gradient) and the intercept method (plot where the line crosses each axis). This page also covers reading intercepts off an equation, and graphing horizontal and vertical lines.

The \(y\)-intercept is where a line crosses the \(y\)-axis; there \(x=0\). The \(x\)-intercept is where it crosses the \(x\)-axis; there \(y=0\). Two points fix a straight line, so a sketch needs just two well-chosen points.

From \(y=mx+c\) you can sketch quickly: plot \((0,c)\), then use the gradient \(m\) as rise over run to step to a second point. A horizontal line \(y=c\) is flat; a vertical line \(x=a\) is upright.

Set the other variable to zero. For the \(x\)-intercept put \(y=0\) and solve for \(x\); for the \(y\)-intercept put \(x=0\) and solve for \(y\).
Sketch from gradient and y-interceptPlot the y intercept, then step across 1 and up 2 for gradient 2 to find a second point. x y (0,-4) 1 2
Gradient-intercept method: plot \((0,c)\), then step across the run and up the rise.
Sketch by the two interceptsPlot the x intercept and y intercept, then draw the line through them. x y (3,0) (0,3)
Intercept method: plot both intercepts and rule the line through them.

To find the intercepts of a line:

\[\text{$y$-intercept: set } x=0 \quad\Rightarrow\quad y=c\]
x=0y=c
\[\text{$x$-intercept: set } y=0 \quad\Rightarrow\quad x=-\dfrac{c}{m}\]
y=0x=-cm
Special lines: \(y=c\) is horizontal through \((0,c)\); \(x=a\) is vertical through \((a,0)\). A vertical line has no \(y\)-intercept unless it is the \(y\)-axis itself.

How to sketch a straight line

  1. Choose a method: from \(y=mx+c\) plot \((0,c)\) and step by the gradient; otherwise find both intercepts.
  2. Find two points: set \(x=0\) for the \(y\)-intercept and \(y=0\) for the \(x\)-intercept, or step off the gradient.
  3. Plot and rule a straight line through the points, extending it and labelling the intercepts.
Example 1 — Sketch from gradient and intercept
Sketch \(y=2x-4\) using its gradient and \(y\)-intercept.
Solution

Read \(m\) and \(c\) from \(y=mx+c\):

\(m\)\(=\)\(2\)
\(c\)\(=\)\(-4\)

Plot the \(y\)-intercept \((0,-4)\).

Find the \(x\)-intercept by setting \(y=0\):

\(0\)\(=\)\(2x-4\)
\(2x\)\(=\)\(4\)
\(x\)\(=\)\(2\)

Line through \((0,-4)\) and \((2,0)\), rising \(2\) for every \(1\) across.

Sketch of y=2x-4Line y equals 2x minus 4 with y intercept 0, negative 4 and x intercept 2, 0. x y (0,-4) (2,0)
(0,-4),(2,0)
Example 2 — Sketch by the intercept method
Sketch \(3x+4y=12\) by finding both intercepts.
Solution

\(x\)-intercept — set \(y=0\):

\(3x+4(0)\)\(=\)\(12\)
\(3x\)\(=\)\(12\)
\(x\)\(=\)\(4\)

\(y\)-intercept — set \(x=0\):

\(3(0)+4y\)\(=\)\(12\)
\(4y\)\(=\)\(12\)
\(y\)\(=\)\(3\)

Line through \((4,0)\) and \((0,3)\).

Intercepts of 3x+4y=12Line 3x plus 4y equals 12 crossing the axes at 4, 0 and 0, 3. x y (4,0) (0,3)
(4,0),(0,3)
Example 3 — Finding an x-intercept
Where does \(y=2x-6\) cross the \(x\)-axis?
Solution

On the \(x\)-axis \(y=0\); substitute and solve:

\(0\)\(=\)\(2x-6\)
\(2x\)\(=\)\(6\)
\(x\)\(=\)\(3\)

\(x\)-intercept at \((3,\,0)\).

x-intercept of y=2x-6Line y equals 2x minus 6 crossing the x axis at 3, 0. x y (3,0) (0,-6)
(3,0)
Example 4 — A vertical line
Find and graph the equation of the line through \((2,\,-5)\) and \((2,\,3)\).
Solution

Compare the coordinates:

\(x_1\)\(=\)\(2\)
\(x_2\)\(=\)\(2\)

Both points have the same \(x\)-value, so the line is vertical. Every point on it has \(x=2\).

Equation: \(x=2\) (a vertical line, gradient undefined).

A vertical lineThe vertical line x equals 2 through the points 2, negative 5 and 2, 3. x y (2,-5) (2,3)
x=2

Common pitfalls

Swapping the intercept conditions. The \(x\)-intercept is where \(y=0\) (not \(x=0\)). Set the other variable to zero for each intercept.
Reading the gradient before rearranging. In a form like \(3x+4y=12\), make \(y\) the subject before reading \(m\) — here \(m=-\tfrac{3}{4}\), not \(3\).
Confusing \(y=2\) with \(x=2\). \(y=c\) is a horizontal line; \(x=a\) is a vertical line. Sketch a quick check point if you are unsure.

Frequently asked questions

How do you sketch a line from y = mx + c?

Plot the \(y\)-intercept \((0,c)\), then use the gradient \(m\) as rise over run to step to a second point, and rule the line.

How do you find the x-intercept and y-intercept?

Set \(y=0\) and solve for \(x\) to get the \(x\)-intercept; set \(x=0\) and solve for \(y\) to get the \(y\)-intercept.

What does the graph of y = 3 look like?

A horizontal line crossing the \(y\)-axis at \(3\); every point has \(y=3\) and the gradient is \(0\).

What does the graph of x = 2 look like?

A vertical line crossing the \(x\)-axis at \(2\); every point has \(x=2\) and the gradient is undefined.

How many points do I need to draw a line?

Two. A straight line is fixed by any two of its points, so plot two (often the intercepts) and rule through them.