Equations of lines
Master the equations of lines for Year 12 Specialist Mathematics in Queensland (QCAA). A straight line is fixed by one point and a direction, and is written three equivalent ways: the vector equation \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\), the parametric equations, and the Cartesian (symmetric) equation, in both two and three dimensions.
You will learn to build the line through two points using \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), convert freely between the three forms, describe line segments with a restricted parameter, and test whether a point lies on a line or whether two lines are parallel — core skills for the 3D vectors topic and everything that follows.
Theory
An equation of a line pins down every point on a straight line using a known point and a direction. In Year 12 Specialist Mathematics (QCAA, Queensland) a line is written three equivalent ways — the vector equation \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\), the parametric equations, and the Cartesian (symmetric) equation — in both two and three dimensions.
A straight line is fixed by one point on it and a direction. If \(\mathbf{a}\) is the position vector of a known point and \(\mathbf{d}\) is any vector along the line, then the position vector \(\mathbf{r}\) of a general point is
\(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\),
where the parameter \(\lambda\) is a scalar that can take any real value. As \(\lambda\) runs over \(\mathbb{R}\), \(\mathbf{r}\) sweeps out the whole line. This is the vector equation of the line.
Given two points \(A\) and \(B\) on the line, the direction is the displacement between them, \(\mathbf{d}=\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\). Any point on the line and any non-zero multiple of the direction may be used, so the equation of a line is not unique.
Writing \(\mathbf{r}=(x,y,z)\) and reading off each coordinate gives the parametric equations \(x=a_1+\lambda d_1\), \(y=a_2+\lambda d_2\), \(z=a_3+\lambda d_3\). Solving each for \(\lambda\) and equating gives the Cartesian (symmetric) equation. Restricting \(\lambda\) to an interval such as \([0,1]\) describes a line segment rather than the full line.
The vector equation of the line through the point with position vector \(\mathbf{a}\) in the direction \(\mathbf{d}\) is:
For a line through two points \(A\) and \(B\), the direction is:
Reading off the coordinates gives the parametric equations:
Eliminating \(\lambda\) gives the Cartesian (symmetric) equation:
How to find the equation of a line
- Find a point and a direction. Use a given point for \(\mathbf{a}\); if two points \(A\), \(B\) are given, take \(\mathbf{d}=\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\).
- Write the vector equation \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\), keeping \(\mathbf{a}\) as the point and \(\mathbf{d}\) as the direction.
- Split into parametric form by reading each coordinate: \(x=a_1+\lambda d_1\), and likewise for \(y\) and \(z\).
- Eliminate \(\lambda\) by solving each parametric equation for \(\lambda\) and equating to get the Cartesian form.
Direction is \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\); then use \(\mathbf{r}=\mathbf{a}+\lambda\overrightarrow{AB}\):
| \(\overrightarrow{AB}\) | \(=\) | \((5-2,\ 3-1,\ 4-0)\) |
| \(=\) | \((3,\,2,\,4)\) | |
| \(\mathbf{r}\) | \(=\) | \(\mathbf{a}+\lambda\overrightarrow{AB}\) |
| \(=\) | \((2\mathbf{i}+\mathbf{j})+\lambda(3\mathbf{i}+2\mathbf{j}+4\mathbf{k})\) |
\(\mathbf{r}=2\mathbf{i}+\mathbf{j}+\lambda(3\mathbf{i}+2\mathbf{j}+4\mathbf{k})\).
Read \(\mathbf{a}=(3,-1,2)\) and \(\mathbf{d}=(1,4,-1)\); each coordinate is \(a_i+\lambda d_i\):
| \(x\) | \(=\) | \(3+\lambda\) |
| \(y\) | \(=\) | \(-1+4\lambda\) |
| \(z\) | \(=\) | \(2-\lambda\) |
Solve each for \(\lambda\) and equate (\(\lambda=\dfrac{x_i-a_i}{d_i}\)):
| \(\lambda\) | \(=\) | \(\dfrac{x-3}{1}\) |
| \(\lambda\) | \(=\) | \(\dfrac{y+1}{4}\) |
| \(\lambda\) | \(=\) | \(\dfrac{z-2}{-1}\) |
| \(\dfrac{x-3}{1}\) | \(=\) | \(\dfrac{y+1}{4}=\dfrac{z-2}{-1}\) |
\(x=3+\lambda,\ y=-1+4\lambda,\ z=2-\lambda\); Cartesian \(\dfrac{x-3}{1}=\dfrac{y+1}{4}=\dfrac{z-2}{-1}\).
Match the \(x\)-coordinate to find \(\lambda\), then check that the SAME \(\lambda\) fits \(y\) and \(z\):
| \(x:\quad 1+2\lambda\) | \(=\) | \(7\) |
| \(2\lambda\) | \(=\) | \(6\) |
| \(\lambda\) | \(=\) | \(3\) |
| \(y:\quad 2-3\) | \(=\) | \(-1\ \checkmark\) |
| \(z:\quad -1+3(3)\) | \(=\) | \(8\ \checkmark\) |
Yes; every coordinate gives \(\lambda=3\), so \((7,-1,8)\) lies on the line.
Parallel lines share a direction, so keep \(\mathbf{d}=(3,-2,5)\) and use \(P\) as the point \(\mathbf{a}\):
| \(\mathbf{d}\) | \(=\) | \((3,-2,5)\) |
| \(\mathbf{a}\) | \(=\) | \((1,0,4)\) |
| \(\mathbf{r}\) | \(=\) | \(\mathbf{a}+\lambda\mathbf{d}\) |
| \(=\) | \((\mathbf{i}+4\mathbf{k})+\lambda(3\mathbf{i}-2\mathbf{j}+5\mathbf{k})\) |
\(\mathbf{r}=\mathbf{i}+4\mathbf{k}+\lambda(3\mathbf{i}-2\mathbf{j}+5\mathbf{k})\).
Common pitfalls
Frequently asked questions
What is the vector equation of a line?
It is \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\), where \(\mathbf{a}\) is the position vector of a known point on the line, \(\mathbf{d}\) is a direction vector along the line, and \(\lambda\) is a scalar parameter that sweeps out every point.
How do you find the equation of a line through two points?
Take the direction \(\mathbf{d}=\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), then write \(\mathbf{r}=\mathbf{a}+\lambda\overrightarrow{AB}\) using either point as \(\mathbf{a}\).
What is the difference between the vector, parametric and Cartesian forms?
They describe the same line. The vector form is \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\); the parametric form lists \(x,y,z\) separately as \(a_i+\lambda d_i\); the Cartesian form eliminates \(\lambda\) to give \(\dfrac{x-a_1}{d_1}=\dfrac{y-a_2}{d_2}=\dfrac{z-a_3}{d_3}\).
How do you tell if a point lies on a line?
Substitute the point and solve one coordinate for \(\lambda\); the point is on the line only if that same \(\lambda\) satisfies every other coordinate.
How do you know if two lines are parallel?
Two lines are parallel when their direction vectors are scalar multiples of one another, that is \(\mathbf{d}_2=k\mathbf{d}_1\) for some scalar \(k\).
Is the equation of a line unique?
No. Any point on the line can be used for \(\mathbf{a}\), and any non-zero multiple of the direction can be used for \(\mathbf{d}\), so a line has many correct vector equations.