1.3: Equations of Lines in 2d

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A line in two dimensions can be specified by giving one point $(x_0,y_0)$ on the line and one vector $\textbf{d}=\left \langle d_x,d_y \right \rangle $ whose direction is parallel to the line.

$twodLine.svg$

If $(x,y)$ is any point on the line then the vector $\left \langle x-x_0,y-y_0 \right \rangle \text{,}$ whose tail is at $(x_0,y_0)$ and whose head is at $(x,y)\text{,}$ must be parallel to $\textbf{d}$ and hence must be a scalar multiple of $\textbf{d}\text{.}$ So

Equation 1.3.1. Parametric Equations

\[\begin{gather*} \left \langle x-x_0,y-y_0 \right \rangle =t \textbf{d} \end{gather*}\]

or, writing out in components,

\[\begin{align*} x-x_0&=t d_x\\ y-y_0&=t d_y \end{align*}\]

These are called the parametric equations of the line, because they contain a free parameter, namely $t\text{.}$ As $t$ varies from $-\infty$ to $\infty\text{,}$ the point $(x_0+td_x,y_0+td_y)$ traverses the entire line.

It is easy to eliminate the parameter $t$ from the equations. Just multiply $x-x_0=t d_x$ by $d_y\text{,}$ multiply $y-y_0=t d_y$ by $d_x$ and subtract to give

\[\begin{gather*} (x-x_0)d_y-(y-y_0)d_x=0 \end{gather*}\]

In the event that $d_x$ and $d_y$ are both nonzero, we can rewrite this as

Equation 1.3.2. Symmetric Equation

\[\begin{gather*} \frac{x-x_0}{d_x}=\frac{y-y_0}{d_y} \end{gather*}\]

This is called the symmetric equation for the line.

A second way to specify a line in two dimensions is to give one point $(x_0,y_0)$ on the line and one vector $\textbf{n}=\left \langle n_x,n_y \right \rangle $ whose direction is perpendicular to that of the line.

$twodLineNormal.svg$

Equation 1.3.3

\[\begin{gather*} \textbf{n}\cdot\left \langle x-x_0,y-y_0 \right \rangle =0 \end{gather*}\]

Writing out in components

\[\begin{gather*} n_x(x-x_0)+n_y(y-y_0)=0\qquad\text{or}\qquad n_xx+n_yy= n_xx_0+n_yy_0 \end{gather*}\]

Observe that the coefficients $n_x,n_y$ of $x$ and $y$ in the equation of the line are the components of a vector $\left \langle n_x,n_y \right \rangle $ perpendicular to the line. This enables us to read off a vector perpendicular to any given line directly from the equation of the line. Such a vector is called a normal vector for the line.

Example 1.3.4

Consider, for example, the line $y=3x+7\text{.}$ To rewrite this equation in the form

\[ n_xx+n_yy= n_xx_0+n_yy_0 \nonumber \]

we have to move terms around so that $x$ and $y$ are on one side of the equation and $7$ is on the other side: $3x-y=-7\text{.}$ Then $n_x$ is the coefficient of $x\text{,}$ namely $3\text{,}$ and $n_y$ is the coefficient of $y\text{,}$ namely $-1\text{.}$ One normal vector for $y=3x+7$ is $\left \langle 3,-1 \right \rangle \text{.}$

Of course, if $\left \langle 3,-1 \right \rangle $ is perpendicular to $y=3x+7\text{,}$ so is $-5\left \langle 3,-1 \right \rangle =\left \langle -15,5 \right \rangle \text{.}$ In fact, if we first multiply the equation $3x-y=-7$ by $-5$ to get $-15x+5y=35$ and then set $n_x$ and $n_y$ to the coefficients of $x$ and $y$ respectively, we get $\textbf{n}=\left \langle -15,5 \right \rangle \text{.}$

Example 1.3.5

In this example, we find the point on the line $y=6-3x$ (call the line $L$) that is closest to the point $(7,5)\text{.}$

We'll start by sketching the line. To do so, we guess two points on $L$ and then draw the line that passes through the two points.

If $(x,y)$ is on $L$ and $x=0\text{,}$ then $y=6\text{.}$ So $(0,6)$ is on $L\text{.}$
If $(x,y)$ is on $L$ and $y=0\text{,}$ then $x=2\text{.}$ So $(2,0)$ is on $L\text{.}$

$closestA.svg$ $closestB.svg$

Denote by $P$ the point on $L$ that is closest to $(7,5)\text{.}$ It is characterized by the property that the line from $(7,5)$ to $P$ is perpendicular to $L\text{.}$ This is the case just because if $Q$ is any other point on $L\text{,}$ then, by Pythagoras, the distance from $(7,5)$ to $Q$ is larger than the distance from $(7,5)$ to $P\text{.}$ See the figure on the right above.

Let's use $N$ to denote the line which passes through $(7,5)$ and which is perpendicular to $L\text{.}$

$closest.svg$

Since $L$ has the equation $3x+y=6\text{,}$ one vector perpendicular to $L\text{,}$ and hence parallel to $N\text{,}$ is $\left \langle 3,1 \right \rangle \text{.}$ So if $(x,y)$ is any point on $N\text{,}$ the vector $\left \langle x-7,y-5 \right \rangle $ must be of the form $t\left \langle 3,1 \right \rangle \text{.}$ So the parametric equations of $N$ are

\[\begin{gather*} \left \langle x-7,y-5 \right \rangle =t\left \langle 3,1 \right \rangle \qquad\text{or}\qquad x=7+3t,\ y=5+t \end{gather*}\]

Now let $(x,y)$ be the coordinates of $P\text{.}$ Since $P$ is on $N\text{,}$ we have $x=7+3t\text{,}$ $y=5+t$ for some $t\text{.}$ Since $P$ is also on $L\text{,}$ we also have $3x+y=6\text{.}$ So

\[\begin{alignat*}{2} & & 3(7+3t)+(5+t)&= 6\\ & \iff\qquad& 10t+26&= 6\\ & \iff\qquad& t&=-2\\ & \implies\qquad& x&= 7+3\times (-2)=1,\ y=5+(-2)=3 \end{alignat*}\]

and $P$ is $(1,3)\text{.}$

Exercises

Stage 1

1

A line in $\mathbb R^2$ has direction $\mathbf d$ and passes through point $\mathbf c\text{.}$

Which of the following gives its parametric equation: $\left \langle x,y \right \rangle =\mathbf c + t\mathbf d \text{,}$ or $\left \langle x,y \right \rangle =\mathbf c - t\mathbf d \text{?}$

2

A line in $\mathbb R^2$ has direction $\mathbf d$ and passes through point $\mathbf c\text{.}$

Which of the following gives its parametric equation: $\left \langle x,y \right \rangle =\mathbf c + t\mathbf d \text{,}$ or $\left \langle x,y \right \rangle =-\mathbf c +t \mathbf d\text{?}$

3

Two points determine a line. Verify that the equations

\[ \left \langle x-1,y-9 \right \rangle=t\left \langle 8,4 \right \rangle \nonumber \]

and

\[ \left \langle x-9,y-13 \right \rangle=t\left \langle 1,\tfrac12 \right \rangle \nonumber \]

describe the same line by finding two different points that lie on both lines.

4

A line in $\mathbb R^2$ has parametric equations

\[ \begin{array}{lcl} x-3&=&9t\\ y-5&=&7t \end{array} \nonumber \]

There are many different ways to write the parametric equations of this line. If we rewrite the equations as

\[ \begin{array}{lcl} x-x_0&=&d_xt\\ y-y_0&=&d_yt \end{array} \nonumber \]

what are all possible values of $\left \langle x_0,y_0 \right \rangle$ and $\left \langle d_x,d_y \right \rangle\text{?}$

Stage 2

5

Find the vector parametric, scalar parametric and symmetric equations for the line containing the given point and with the given direction.

point $(1,2)\text{,}$ direction $\left \langle 3,2 \right \rangle $
point $(5,4)\text{,}$ direction $\left \langle 2,-1 \right \rangle $
point $(-1,3)\text{,}$ direction $\left \langle -1,2 \right \rangle $

6

Find the vector parametric, scalar parametric and symmetric equations for the line containing the given point and with the given normal.

point $(1,2)\text{,}$ normal $\left \langle 3,2 \right \rangle $
point $(5,4)\text{,}$ normal $\left \langle 2,-1 \right \rangle $
point $(-1,3)\text{,}$ normal $\left \langle -1,2 \right \rangle $

7

Use a projection to find the distance from the point $(-2,3)$ to the line $3x-4y=-4\text{.}$

8

Let $\textbf{a} \text{,}$ $\textbf{b} $ and $\textbf{c} $ be the vertices of a triangle. By definition, a median of a triangle is a straight line that passes through a vertex of the triangle and through the midpoint of the opposite side.

Find the parametric equations of the three medians.
Do the three medians meet at a common point? If so, which point?

9

Let $C$ be the circle of radius 1 centred at $(2,1)\text{.}$ Find an equation for the line tangent to $C$ at the point $\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)\text{.}$

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