1.3: Equations of Lines in 2d

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.

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.

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 perpendicular to \(\textbf{n}\) so that

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 \(L\text{.}\)

            

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 \(L\text{.}\)

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.

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

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

1. Find the parametric equations of the three medians.
2. 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 \(\left(\frac{5}{2},1+\frac{\sqrt3}{2}\right)\text{.}\)