Equation of a line

A GPS gives you a starting point and a heading. Together they determine a straight road, with no further information. That is exactly how we describe a line in coordinates: a point on the line and a direction. This lesson turns that idea into equations in three useful forms.

Direction vector and normal vector

A direction vector $\vec{u}=(a, b)\neq\vec{0}$ of a line is a vector parallel to the line. A normal vector $\vec{n}=(A, B)\neq\vec{0}$ is perpendicular to the line. Since $\vec{u}\perp\vec{n}$, rotating $\vec{u}=(a, b)$ gives $\vec{n}=(-b, a)$ (or $(b, -a)$). Any non-zero multiple of $\vec{u}$ (or $\vec{n}$) works too.

Three forms of the equation

In the general form the coefficients $(A, B)$ are a normal vector of the line. If $B\neq 0$, the slope is $k=-\frac{A}{B}$.

A line through two points

For $A(x_A, y_A)$ and $B(x_B, y_B)$ take $\overrightarrow{AB}$ as a direction vector, then a normal vector, then the general form.

Converting and sketching

Parallel and perpendicular lines

A line parallel to $Ax+By+C=0$ is $Ax+By+C'=0$ (same $\vec{n}$). A line perpendicular to it has $\vec{n}'=\vec{u}=(B, -A)$ as its normal vector, i.e. it is $Bx-Ay+C''=0$.

Key terms

Summary

• A line is fixed by a point and a direction vector $\vec{u}$ or a normal vector $\vec{n}$.
• Parametric: $x=x_0+at$, $y=y_0+bt$. General: $A(x-x_0)+B(y-y_0)=0$ with $\vec{n}=(A, B)$.
• Slope form $y=kx+m$ with $k=-\frac{A}{B}$ for $Ax+By+C=0$ and $B\neq0$.
• Parallel lines share $\vec{n}$; perpendicular lines swap $(A, B)\to(B, -A)$.

Practice

Show solutions

1. $(x-2)-3(y-1)=0$, so $x-3y+1=0$.
2. $\begin{cases}x=1+2t\\ y=1+t\end{cases}$. $\vec{n}=(1, -2)$: $(x-1)-2(y-1)=0$, so $x-2y+1=0$.
3. $\overrightarrow{AB}=(4, -4)$, so $\vec{u}=(1, -1)$ and $\vec{n}=(1, 1)$. $(x+1)+(y-2)=0$, so $x+y-1=0$.
4. $y=2x+4$: slope $2$. For $y=0$, $x=-2$; for $x=0$, $y=4$. Intercepts $(-2, 0)$ and $(0, 4)$.
5. $\vec{u}=(-1, 2)$, $\vec{n}=(2, 1)$, point $(2, 3)$: $2(x-2)+(y-3)=0$, so $2x+y-7=0$.
6. $2x-y+C=0$ through $P$: $2+2+C=0$, so $C=-4$. Line: $2x-y-4=0$.
7. $\vec{n}'=(1, 2)$: $(x-1)+2(y+2)=0$, so $x+2y+3=0$. Check: $1-4+3=0$ ✓.
8. $\frac{x}{4}+\frac{y}{-3}=1$; multiplying by $12$: $3x-4y=12$, so $3x-4y-12=0$.
9. $M(3, 2)$, $\vec{n}=\overrightarrow{AB}=(4, 2)$: $4(x-3)+2(y-2)=0$, so $2x+y-8=0$. Check $M$: $6+2-8=0$ ✓.
10. $M$ is the midpoint: $\frac{a+0}{2}=2$ and $\frac{0+b}{2}=3$, so $a=4$, $b=6$. The line is $\frac{x}{4}+\frac{y}{6}=1$, i.e. $3x+2y-12=0$.