Drawing a Straight Line

Given an equation like \(y = 2x - 1\), you can turn it into a picture. Choose a few \(x\)-values, work out the matching \(y\)-values in a table, plot the points, and join them with a ruler.

MYP 4Standard MathsCoordinate GeometryCriteria A · C · D~10 min read

Why a table of values

An equation such as \(y = 2x - 1\) is a rule: give it an \(x\), and it hands back a \(y\). Each \(x\) and \(y\) pair is a coordinate you can plot. A table of values is just a tidy way of collecting several of those pairs before you draw.

Table of values
A short list of chosen \(x\)-values with the \(y\)-values the equation gives for each one. Every column is one point \((x, y)\) on the line.

Building the table

Pick three or four small \(x\)-values that fit your grid, such as \(-1, 0, 1, 2\). Substitute each one into the equation to find \(y\). Working with \(y = 2x - 1\):

\(x\)\(-1\)\(0\)\(1\)\(2\)
\(y\)\(-3\)\(-1\)\(1\)\(3\)

Each column now reads as a point: \((-1, -3)\), \((0, -1)\), \((1, 1)\) and \((2, 3)\).

Always plot at least three points

Two points are enough to define a line, but a third acts as a checker. If it does not sit on the same straight path as the other two, you have an arithmetic slip to fix before drawing.

Plotting and joining

Mark each point on the grid, then join them with a single straight ruled line and extend it a little past the outer points. Because the equation is linear, the points must line up perfectly. Label the line with its equation.

Worked example

Worked example

Draw the line \(y = 3x + 1\) for \(x\) from \(-1\) to \(2\).

1
Choose \(x = -1, 0, 1, 2\) and substitute each into \(y = 3x + 1\).
2
Work out the values: \(3(-1)+1 = -2\), \(3(0)+1 = 1\), \(3(1)+1 = 4\), \(3(2)+1 = 7\).
3
The points are \((-1, -2)\), \((0, 1)\), \((1, 4)\) and \((2, 7)\). Plot them and join with a ruler.
A straight line through \((0, 1)\) rising 3 units for every 1 unit right

Check yourself

1. For \(y = 2x + 1\), what is \(y\) when \(x = 4\)? +

Substitute: \(y = 2(4) + 1 = 8 + 1\). So \(y = \mathbf{9}\), giving the point \((4, 9)\).

2. In a table for \(y = -x + 5\), what \(y\) goes with \(x = 2\)? +

Substitute: \(y = -(2) + 5 = 3\). The point is \(\mathbf{(2, 3)}\).

3. You plot \(y = x - 3\) and your three points do not line up. What has gone wrong? +

A linear equation always gives points on one straight line, so a point out of line means an arithmetic error in the table. Recheck each \(y\)-value before drawing.


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