Finding an Equation of a Line
Every question with a fully worked solution.
- Finding an Equation of a Line - Video - The Equation of a Line Watch
Theory
This Year 10 maths page revises how to read and write the equation of a straight line from Years 7 to 9: the gradient and \(y\)-intercept in \(y = mx + c\), changing to and from general form, finding intercepts, the line through two points, through a point with a given gradient or parallel or perpendicular to another line, reading a line from a graph, and using a line as a model of a real situation.
Every straight line (except a vertical one) has an equation \(y = mx + c\). This is the gradient–intercept form: \(m\) is the gradient and \(c\) is the \(y\)-intercept, where the line crosses the \(y\)-axis. Some books write \(y = mx + b\); \(b\) means the same as \(c\).
The general form is \(ax + by + c = 0\), with every term on one side and whole-number coefficients, usually with \(a\) positive. For example \(y = 3x - 4\) becomes \(3x - y - 4 = 0\).
The \(x\)-intercept is where the line crosses the \(x\)-axis (there \(y = 0\)). The \(y\)-intercept is where it crosses the \(y\)-axis (there \(x = 0\)).
A point lies on a line if its coordinates make the equation true.
In a real-life model, the gradient is the rate of change (for example litres per hour) and the \(y\)-intercept is the starting value.
The three forms of a straight line:
How to find the equation of a line
- Find the gradient \(m\): it may be given, come from two points, or come from a parallel or perpendicular line.
- Use a point. Either substitute the point into \(y = mx + c\) and solve for \(c\), or use \(y - y_1 = m(x - x_1)\).
- Write the equation in the form asked for. For general form, clear fractions and move every term to one side.
- Check by substituting a second point into your equation.
How to find the intercepts
- \(x\)-intercept: put \(y = 0\) and solve for \(x\).
- \(y\)-intercept: put \(x = 0\) and solve for \(y\).
- Write both as points, such as \((5, 0)\) and \((0, -3)\).
(i) \(y = 5 - 4x\)
(ii) \(5x + 2y - 8 = 0\). Also find the \(x\)-intercept of line (ii).
(i) Write the \(x\) term first: \(y = -4x + 5\). So \(\boldsymbol{m = -4}\) and \(\boldsymbol{c = 5}\).
(ii) Make \(y\) the subject:
| \(2y\) | \(=\) | \(-5x + 8\) |
| \(y\) | \(=\) | \(-\dfrac{5}{2}x + 4\) |
So \(\boldsymbol{m = -\dfrac{5}{2}}\) and \(\boldsymbol{c = 4}\). For the \(x\)-intercept, put \(y = 0\):
| \(5x - 8\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{8}{5} = 1.6\) |
The \(x\)-intercept is \(\boldsymbol{(1.6, 0)}\).
Find the gradient first.
| \(m\) | \(=\) | \(\dfrac{3 - (-1)}{4 - (-2)} = \dfrac{4}{6} = \dfrac{2}{3}\) |
Use \(y - y_1 = m(x - x_1)\) with the point \((4, 3)\):
| \(y - 3\) | \(=\) | \(\dfrac{2}{3}(x - 4)\) |
| \(y\) | \(=\) | \(\dfrac{2}{3}x - \dfrac{8}{3} + 3\) |
| \(y\) | \(=\) | \(\boldsymbol{\dfrac{2}{3}x + \dfrac{1}{3}}\) |
For general form, multiply by \(3\) and move every term to one side:
| \(3y\) | \(=\) | \(2x + 1\) |
| \(0\) | \(=\) | \(\boldsymbol{2x - 3y + 1}\) |
Check with the other point \((-2, -1)\): \(-4 + 3 + 1 = 0\) ✓
(i) Write an equation for the charge \(B\%\) after \(t\) hours.
(ii) What was the charge at the start?
(iii) After how many hours is it flat?
(i) The points \((2, 74)\) and \((5, 47)\) lie on a straight line. The gradient is the rate of change.
| \(m\) | \(=\) | \(\dfrac{47 - 74}{5 - 2} = \dfrac{-27}{3} = -9\) |
| \(74\) | \(=\) | \(-9(2) + c\) |
| \(c\) | \(=\) | \(92\) |
So \(\boldsymbol{B = -9t + 92}\). The battery loses \(9\%\) each hour.
(ii) At the start \(t = 0\), so \(B = 92\): the charge was \(\boldsymbol{92\%}\).
(iii) Put \(B = 0\):
| \(-9t + 92\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(\dfrac{92}{9} \approx \boldsymbol{10.2~\text{hours}}\) |
(i) Find the equation of the line through \((-2, 6)\) that is perpendicular to \(y = 2x - 7\).
(ii) Does the point \((10, 0)\) lie on this line?
(i) The given gradient is \(2\), so the perpendicular gradient is \(-\dfrac{1}{2}\). Substitute the point into \(y = mx + c\) to find \(c\).
| \(6\) | \(=\) | \(-\dfrac{1}{2}(-2) + c\) |
| \(6\) | \(=\) | \(1 + c\) |
| \(c\) | \(=\) | \(5\) |
The line is \(\boldsymbol{y = -\dfrac{1}{2}x + 5}\).
(ii) Substitute \(x = 10\):
| \(y\) | \(=\) | \(-\dfrac{1}{2}(10) + 5 = 0\) |
This matches the \(y\)-value, so yes, \((10, 0)\) lies on the line.
Common pitfalls
Frequently asked questions
What does y = mx + c mean?
It is the equation of a straight line. \(m\) is the gradient (how steep the line is) and \(c\) is the \(y\)-intercept (where it crosses the \(y\)-axis).
How do you find the equation of a line through two points?
Find the gradient with \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). Then substitute one point into \(y = mx + c\) to find \(c\), and check with the other point.
What is the general form of a straight line?
It is \(ax + by + c = 0\): every term on one side, equal to zero, with whole-number coefficients.
How do you find the x-intercept of a line?
Put \(y = 0\) in the equation and solve for \(x\). The \(x\)-intercept is the point \((x, 0)\).
How do you check whether a point is on a line?
Substitute the \(x\)-value of the point into the equation. If you get the \(y\)-value of the point, the point is on the line.
How do you find the equation of a line from a graph?
Read the \(y\)-intercept, then use a second exact grid point to find the gradient as rise over run. Write \(y = mx + c\) and check with another point on the line.