Graphing Inequalities on the Cartesian Plane
Theory
A linear inequality in \(x\) and \(y\), such as \(y \lt 2x + 1\), describes a region of the number plane: all the points on one side of a boundary line. This Year 10A maths page (Victorian Curriculum 10A) shows how to test whether a point is in a region, draw the boundary solid or dashed, shade the correct side using a test point, write the inequality for a shaded region, and describe regions made by two or more inequalities. It also finds the corners and areas of regions, and spots regions that never overlap. In the Victorian Curriculum this is part of VC2M10A08 (graphing regions); in the Australian Curriculum it interprets inequalities graphically. Curriculum: VC2M10A08 / AC9M10A02.
A linear inequality in two variables, such as \(y \lt 2x + 1\) or \(3x + 2y \ge 6\), is true for some points \((x, y)\) and false for others. All the points that make it true form a region of the number plane: everything on one side of a straight line.
The boundary line comes from replacing the inequality sign with \(=\). Draw it solid for \(\le\) or \(\ge\), because points on the line are included, and dashed for \(\lt\) or \(\gt\), because they are not.
A test point decides which side to shade. Substitute a point that is not on the line, usually the origin \(O(0, 0)\). If the inequality is true, shade the side containing the test point; if it is false, shade the other side.
When \(y\) is the subject, \(y \gt mx + c\) is the region above the line and \(y \lt mx + c\) is the region below it. For a vertical boundary, \(x \gt a\) is the region to the right of the line \(x = a\).
A region described by two or more inequalities is their overlap: the points that make every inequality true at the same time.
| Sign | Boundary line | Points on the line |
|---|---|---|
| \(\lt\) or \(\gt\) | dashed | not in the region |
| \(\le\) or \(\ge\) | solid | in the region |
| Inequality | Region |
|---|---|
| \(y \gt mx + c\) | above the line |
| \(y \lt mx + c\) | below the line |
| \(x \gt a\) | right of the vertical line \(x = a\) |
| \(y \lt b\) | below the horizontal line \(y = b\) |
How to sketch a region
- Boundary: replace the inequality sign with \(=\) and find two points on the line, such as the intercepts.
- Solid or dashed: solid for \(\le\) or \(\ge\), dashed for \(\lt\) or \(\gt\).
- Test a point not on the line, usually \(O(0, 0)\), in the original inequality.
- Shade the side containing the test point if it gave a true statement, otherwise the other side.
- Label the boundary with its equation.
How to write the inequality for a shaded region
- Find the equation of the boundary line from two points on it.
- Choose the sign type: \(\lt\) or \(\gt\) for dashed, \(\le\) or \(\ge\) for solid.
- Substitute a shaded point: the direction that makes it true is the one to use.
Two or more inequalities
- Draw each boundary and decide which side of each one works.
- Shade only the overlap, where every inequality is true.
- Check one point inside the region in every inequality. For a worded problem, also include conditions such as \(x \ge 0\) and \(y \ge 0\) when \(x\) and \(y\) count things.
Substitute each point into the inequality and check whether it is true.
| \((1, 4):\ 4\) | \(\le\) | \(2(1) + 1 = 3 \quad \text{false}\) |
| \((0, 1):\ 1\) | \(\le\) | \(2(0) + 1 = 1 \quad \text{true}\) |
| \((3, 0):\ 0\) | \(\le\) | \(2(3) + 1 = 7 \quad \text{true}\) |
| \((-2, 1):\ 1\) | \(\le\) | \(2(-2) + 1 = -3 \quad \text{false}\) |
The points in the region are \(\mathbf{(0, 1)}\) and \(\mathbf{(3, 0)}\). The point \((0, 1)\) lies on the boundary line; it is included because the sign is \(\le\).
(i) Find the intercepts of the boundary line.
(ii) Is the boundary solid or dashed?
(iii) Use a test point to decide which side to shade.
(i) The boundary is \(4x + y = 4\).
| \(x = 0:\ y\) | \(=\) | \(4 \quad (0, 4)\) |
| \(y = 0:\ 4x\) | \(=\) | \(4,\ x = 1 \quad (1, 0)\) |
(ii) The sign is \(\gt\), so the line is dashed.
(iii) Test \(O(0, 0)\):
| \(4(0) + 0\) | \(\gt\) | \(4\) |
| \(0\) | \(\gt\) | \(4 \quad \text{false}\) |
The origin is not in the region, so shade the side of the line away from \(O\).
Find the equation of the boundary from \((0, -1)\) and \((2, 3)\).
| \(m\) | \(=\) | \(\dfrac{3 - (-1)}{2 - 0} = \dfrac{4}{2} = 2\) |
| \(c\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(2x - 1\) |
The line is solid, so the sign is \(\le\) or \(\ge\). The origin is shaded, so test it:
| \(0\) | \(\;?\;\) | \(2(0) - 1 = -1\) |
| \(0\) | \(\ge\) | \(-1 \quad \text{true}\) |
The inequality is \(\boldsymbol{y \ge 2x - 1}\). The shading is above the line, which agrees with \(y \ge \ldots\).
(i) Sketch the region and find the coordinates of its corners.
(ii) Find its area.
(i) All three boundaries are solid. \(y \ge 0\) is on or above the \(x\)-axis, and \(O\) satisfies the other two (\(0 \le 4\) and \(0 \le 7\)), so the region is the triangle that contains \(O\).
| \(2x + 4 = 0\) | \(\Rightarrow\) | \((-2, 0)\) |
| \(-x + 7 = 0\) | \(\Rightarrow\) | \((7, 0)\) |
| \(2x + 4\) | \(=\) | \(-x + 7\) |
| \(3x\) | \(=\) | \(3,\ \ x = 1,\ \ y = 6\) |
The corners are \(\mathbf{(-2, 0)}\), \(\mathbf{(7, 0)}\) and \(\mathbf{(1, 6)}\).
(ii) The base runs along the \(x\)-axis from \(-2\) to \(7\), and the height is the \(y\)-coordinate of \((1, 6)\).
| \(b\) | \(=\) | \(7 - (-2) = 9\) |
| \(h\) | \(=\) | \(6\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 9 \times 6 = \mathbf{27}\ \text{units}^2\) |
Check with a point inside, such as \((1, 2)\): \(2 \ge 0\), \(2 \le 6\) and \(2 \le 6\) ✓
Common pitfalls
Frequently asked questions
How do you know which side of the line to shade?
Substitute a test point that is not on the line, usually the origin. If it makes the inequality true, shade the side it is on. If it makes it false, shade the other side.
When is the boundary line dashed?
When the inequality is strict, \(\lt\) or \(\gt\). Points on the line do not satisfy it, so the line is drawn dashed. For \(\le\) or \(\ge\) the line is solid.
What does y > 2x + 1 look like on a graph?
It is the region above the line \(y = 2x + 1\), which is drawn dashed. The line crosses the \(y\)-axis at \(1\) and has gradient \(2\).
How do you graph two inequalities on the same number plane?
Draw both boundary lines, decide which side of each one works, and shade only the overlap. Every point in the final region makes both inequalities true.
Can the origin always be used as a test point?
Only if it is not on the boundary line. If the line passes through the origin, use another point such as \((1, 0)\).