Lines with One Intercept
Every question with a fully worked solution.
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Theory
Most straight lines cross both axes at different points, but some do not. A horizontal line \(y = b\) crosses only the \(y\)-axis, a vertical line \(x = a\) crosses only the \(x\)-axis, and a line through the origin, \(y = mx\), has both intercepts at \((0, 0)\). This Year 9 maths page shows how to write and sketch these lines, find the gradient of a line through the origin, and work with shapes they enclose.
A horizontal line has an equation \(y = b\). Every point on it has the same \(y\)-coordinate, \(b\). It is parallel to the \(x\)-axis, crosses the \(y\)-axis at \((0, b)\) and never meets the \(x\)-axis (unless it is the \(x\)-axis, \(y = 0\)). Its gradient is \(0\).
A vertical line has an equation \(x = a\). Every point on it has the same \(x\)-coordinate, \(a\). It is parallel to the \(y\)-axis and crosses the \(x\)-axis at \((a, 0)\). Its gradient is undefined. The \(y\)-axis is \(x = 0\).
A line through the origin has an equation \(y = mx\), with no number added on. Both of its intercepts are the origin \((0, 0)\), so the intercept method gives only one point. Find a second point by choosing another \(x\)-value.
In \(y = mx\), \(m\) is the gradient. For any point \((x, y)\) on the line other than the origin, \(m = \dfrac{y}{x}\).
| The line | Equation | Crosses |
|---|---|---|
| horizontal, through \((a, b)\) | \(y = b\) | \(y\)-axis at \((0, b)\) |
| vertical, through \((a, b)\) | \(x = a\) | \(x\)-axis at \((a, 0)\) |
| through the origin, gradient \(m\) | \(y = mx\) | both axes at \((0, 0)\) |
| the \(x\)-axis | \(y = 0\) | (it is an axis) |
| the \(y\)-axis | \(x = 0\) | (it is an axis) |
How to sketch a line with one intercept
- Rearrange the equation to \(x = a\), \(y = b\) or \(y = mx\).
- \(x = a\): rule a vertical line through \(a\) on the \(x\)-axis.
- \(y = b\): rule a horizontal line through \(b\) on the \(y\)-axis.
- \(y = mx\): plot the origin, then substitute another \(x\)-value (such as \(x = 1\) or \(x = 2\)) to get a second point. Rule the line through both.
- Label each line with its equation.
How to write the equation
- Parallel to the \(x\)-axis (horizontal): use the point's \(y\)-coordinate, \(y = b\).
- Parallel to the \(y\)-axis (vertical): use the point's \(x\)-coordinate, \(x = a\).
- Through the origin and \((x, y)\): work out \(m = \dfrac{y}{x}\) and write \(y = mx\).
(i) Write the equation of the line through the point that is parallel to the \(y\)-axis.
(ii) Write the equation of the line through the point that is parallel to the \(x\)-axis.
(iii) Sketch both lines.
(i) A line parallel to the \(y\)-axis is vertical. Every point on it has the same \(x\)-coordinate as \((-4, 1)\).
| \(x\) | \(=\) | \(\mathbf{-4}\) |
(ii) A line parallel to the \(x\)-axis is horizontal. Every point on it has the same \(y\)-coordinate.
| \(y\) | \(=\) | \(\mathbf{1}\) |
(iii) The two lines cross at \((-4, 1)\), the point they were built from.
(i) Find its equation.
(ii) Find \(y\) when \(x = 9\).
(iii) The point \((k, -10)\) is on the line. Find \(k\).
(i) The line passes through the origin, so it has the form \(y = mx\). Use the point to find \(m\).
| \(m\) | \(=\) | \(\dfrac{y}{x} = \dfrac{4}{6} = \dfrac{2}{3}\) |
| \(y\) | \(=\) | \(\boldsymbol{\dfrac{2}{3}x}\) |
(ii)
| \(y\) | \(=\) | \(\dfrac{2}{3} \times 9 = \mathbf{6}\) |
(iii) Substitute \(y = -10\) and solve.
| \(-10\) | \(=\) | \(\dfrac{2}{3}k\) |
| \(-30\) | \(=\) | \(2k\) |
| \(k\) | \(=\) | \(\mathbf{-15}\) |
(i) \(y + 3 = 0\)
(ii) \(x + 2y = 0\)
and find where the two lines meet.
(i) Subtract \(3\) from both sides: \(y = -3\), a horizontal line.
(ii) Make \(y\) the subject.
| \(2y\) | \(=\) | \(-x\) |
| \(y\) | \(=\) | \(-\dfrac{1}{2}x\) |
Putting \(x = 0\) or \(y = 0\) gives only \((0, 0)\), so choose \(x = 2\): \(y = -1\). Rule the line through \((0, 0)\) and \((2, -1)\).
The lines meet where \(y = -3\) on the second line.
| \(-\dfrac{1}{2}x\) | \(=\) | \(-3\) |
| \(x\) | \(=\) | \(6\) |
They meet at \(\mathbf{(6, -3)}\).
(i) Find the coordinates of its three vertices.
(ii) Find its area.
(i) Each vertex is where two of the lines meet. Substitute one equation into the other.
| \(y = 5,\ x = -1\) | \(:\) | \((-1, 5)\) |
| \(y = 5,\ y = x\) | \(:\) | \((5, 5)\) |
| \(x = -1,\ y = x\) | \(:\) | \((-1, -1)\) |
(ii) The sides along \(y = 5\) and \(x = -1\) meet at a right angle at \((-1, 5)\).
| \(\text{base}\) | \(=\) | \(5 - (-1) = 6\) |
| \(\text{height}\) | \(=\) | \(5 - (-1) = 6\) |
| \(\text{area}\) | \(=\) | \(\dfrac{1}{2} \times 6 \times 6\) |
| \(\) | \(=\) | \(\mathbf{18} \text{ square units}\) |
Common pitfalls
Frequently asked questions
What does the graph of y = 5 look like?
A horizontal line through \(5\) on the \(y\)-axis. Every point on it has \(y\)-coordinate \(5\), whatever its \(x\)-coordinate.
Why is x = 2 a vertical line?
Every point on it has \(x\)-coordinate \(2\), so the points stack straight up and down above and below \(2\) on the \(x\)-axis.
How do you graph a line like y = 4x?
It passes through the origin. Substitute another \(x\)-value, such as \(x = 1\), to get a second point \((1, 4)\), then rule the line through both points.
How do you find the equation of a line through the origin and a point?
The equation is \(y = mx\). Divide the point's \(y\)-coordinate by its \(x\)-coordinate to get \(m\).
What is the equation of the x-axis?
\(y = 0\), because every point on the \(x\)-axis has a \(y\)-coordinate of \(0\). The \(y\)-axis is \(x = 0\).