2.1: Lines in ℝ² - Math 124 Course Notes

import numpy as np
import plotly.graph_objects as go
from pathlib import Path
import sys

_notes_root = next(p for p in [Path.cwd(), *Path.cwd().parents] if (p / 'myst.yml').exists())
if str(_notes_root) not in sys.path:
    sys.path.insert(0, str(_notes_root))
from plot_style import style_plotly

BLUE, ORANGE, PINK = '#3d81f6', 'orange', '#d81a60'
ANNOTATION_SIZE = 24


def base2(xrange=(-5, 6), yrange=(-5, 5), width=700, height=580, tick=1):
    fig = style_plotly(go.Figure(), renderer='png')
    fig.update_layout(width=width, height=height, showlegend=False,
                      font=dict(size=17), margin=dict(l=55, r=30, t=30, b=50))
    fig.update_xaxes(title='x', range=xrange, dtick=tick, zeroline=True, zerolinewidth=1.5,
                     constrain='domain', scaleanchor='y', scaleratio=1)
    fig.update_yaxes(title='y', range=yrange, dtick=tick, zeroline=True, zerolinewidth=1.5,
                     constrain='domain')
    return fig


def line2(fig, start, end, color=BLUE, dash='solid', width=2):
    fig.add_trace(go.Scatter(x=[start[0], end[0]], y=[start[1], end[1]], mode='lines',
                            line=dict(color=color, width=width, dash=dash),
                            hoverinfo='skip', showlegend=False))


def label2(fig, xy, label, dx=0, dy=0, color='black'):
    # MathJax needs its own size command when exporting math labels to PNG.
    if label.startswith('$') and label.endswith('$'):
        label = r'$\Large ' + label[1:]
    fig.add_annotation(x=float(xy[0])+dx, y=float(xy[1])+dy, text=label, showarrow=False,
                       font=dict(family='Palatino', color=color, size=ANNOTATION_SIZE))


def point2(fig, xy, label=None, dx=0.35, dy=0.35):
    fig.add_trace(go.Scatter(x=[xy[0]], y=[xy[1]], mode='markers',
                            marker=dict(color='black', size=7), hoverinfo='skip'))
    if label: label2(fig, xy, '$' + label.replace('−', '-') + '$', dx, dy)


def vec2(fig, end, label=None, color=BLUE, start=(0, 0), dx=0.3, dy=0.3, width=4):
    # Match Chapter 1.4: a colored shaft, arrowhead, and midpoint math label.
    line2(fig, start, end, color=color, width=width)
    fig.add_annotation(x=float(end[0]), y=float(end[1]), ax=float(start[0]), ay=float(start[1]),
                       xref='x', yref='y', axref='x', ayref='y', showarrow=True,
                       text='', arrowhead=3, arrowsize=1.2, arrowwidth=3, arrowcolor=color)
    if label:
        midpoint = (np.asarray(start, dtype=float) + np.asarray(end, dtype=float)) / 2
        label2(fig, midpoint, label, dx, dy, color=color)


def square2(fig, corner, along, perp, size=0.3):
    corner=np.asarray(corner,dtype=float)
    u=np.asarray(along,dtype=float); u=u/np.linalg.norm(u)*size
    v=np.asarray(perp,dtype=float); v=v/np.linalg.norm(v)*size
    line2(fig,corner+u,corner+u+v,'gray',width=1.5)
    line2(fig,corner+v,corner+u+v,'gray',width=1.5)


def base3():
    fig=style_plotly(go.Figure(),renderer='plotly_mimetype')
    axis=dict(range=[-6,7],dtick=2,showbackground=True,showspikes=False)
    fig.update_layout(width=760,height=620,showlegend=False,font=dict(size=16),
                      margin=dict(l=0,r=0,t=15,b=0),
                      scene=dict(xaxis=dict(title='x',**axis),yaxis=dict(title='y',**axis),
                                 zaxis=dict(title='z',**axis),aspectmode='cube',
                                 camera=dict(eye=dict(x=1.6,y=-2.1,z=1.3))))
    for direction in np.eye(3):
        line3(fig,-6*direction,7*direction,'#9ca3af',width=2)
    return fig


def line3(fig,start,end,color=BLUE,width=5,dash='solid'):
    fig.add_trace(go.Scatter3d(x=[start[0],end[0]],y=[start[1],end[1]],z=[start[2],end[2]],
                              mode='lines',line=dict(color=color,width=width,dash=dash),
                              hoverinfo='skip',showlegend=False))


def vec3(fig,end,label,color=BLUE,start=(0,0,0),offset=(0.25,0.25,0.35)):
    start,end=np.asarray(start,float),np.asarray(end,float)
    direction=(end-start)/np.linalg.norm(end-start)
    line3(fig,start,end-0.15*direction,color,width=7)
    fig.add_trace(go.Cone(x=[end[0]],y=[end[1]],z=[end[2]],u=[direction[0]],v=[direction[1]],w=[direction[2]],
                         anchor='tip',sizemode='absolute',sizeref=0.45,colorscale=[[0,color],[1,color]],
                         showscale=False,hoverinfo='skip'))
    pos=end+offset
    fig.add_trace(go.Scatter3d(x=[pos[0]],y=[pos[1]],z=[pos[2]],mode='text',text=[label],
                              textfont=dict(color='black',size=16),hoverinfo='skip'))


def plane3(fig,normal,color=BLUE,opacity=0.24):
    # Solve n_x*x + n_y*y + n_z*z = 0 for y on a rectangular x,z mesh.
    x,z=np.meshgrid(np.linspace(-5,6,2),np.linspace(-5,6,2))
    n=np.asarray(normal,float)
    y=-(n[0]*x+n[2]*z)/n[1]
    fig.add_trace(go.Surface(x=x.tolist(),y=y.tolist(),z=z.tolist(),
                            colorscale=[[0,color],[1,color]],opacity=opacity,
                            showscale=False,hoverinfo='skip'))
    corners=np.array([[x[0,0],y[0,0],z[0,0]],[x[0,1],y[0,1],z[0,1]],
                      [x[1,1],y[1,1],z[1,1]],[x[1,0],y[1,0],z[1,0]]])
    for i in range(4): line3(fig,corners[i],corners[(i+1)%4],color,width=2)


def parametric_line_plot(start=(0, 0)):
    """A draggable Plotly slider for exploring a parametric line."""
    start = np.asarray(start, dtype=float)
    affine = bool(np.any(start))
    color = ORANGE if affine else BLUE
    values = [round(float(t), 2) for t in np.linspace(-1.5, 1.5, 61)]
    fig = base2((-7.5, 8.5), (-6.5, 6.5), width=600, height=620, tick=1)
    fig.update_layout(margin=dict(l=55, r=30, t=95, b=180))
    line2(fig, (-8, -6), (9, 6.75), BLUE, width=2)
    if affine:
        line2(fig, (-8, -7.75), (9, 5), ORANGE, width=2)
    point2(fig, start)
    moving_index = len(fig.data)

    def state(t):
        end = start + t * np.array([4, 3])
        trace = go.Scatter(
            x=[float(start[0]), float(end[0])],
            y=[float(start[1]), float(end[1])],
            mode='lines+markers', line=dict(color=color, width=4),
            marker=dict(color=color, size=[0, 9]), hoverinfo='skip')
        annotations = [dict(
            x=float(end[0]), y=float(end[1]), ax=float(start[0]), ay=float(start[1]),
            xref='x', yref='y', axref='x', ayref='y', text='', showarrow=t != 0,
            arrowhead=3, arrowsize=1.2, arrowwidth=3, arrowcolor=color), dict(
            x=float(end[0]), y=float(end[1]), text=f'({end[0]:g}, {end[1]:g})',
            showarrow=False, yshift=45, font=dict(color=color, size=18)), dict(
            x=0, y=1.1, xref='paper', yref='paper', xanchor='left',
            text=('3x − 4y = 7<br>(x, y) = (1 + 4t, −1 + 3t)'
                  if affine else '3x − 4y = 0     |     (x, y) = (4t, 3t)')
                 + f'     |     t = {t:g}',
            showarrow=False, font=dict(color=color, size=20 if affine else 16))]
        if affine:
            annotations.append(dict(x=1, y=-1, text='Starting point (1, −1)',
                                    showarrow=False, xshift=35, yshift=-24, font=dict(size=16)))
        return dict(data=[trace.to_plotly_json()], traces=[moving_index],
                    layout=dict(annotations=annotations))

    initial = state(1)
    fig.add_trace(initial['data'][0])
    fig.update_layout(**initial['layout'])
    # Each slider step selects the corresponding vector.
    fig.frames = [go.Frame(name=f'{t:g}', **state(t)) for t in values]
    instant = dict(mode='immediate', frame=dict(duration=0, redraw=True),
                   transition=dict(duration=0))
    fig.update_layout(
        sliders=[dict(active=values.index(1.0), x=0.15, len=0.7, y=-0.08,
                      currentvalue=dict(prefix='t = ', font=dict(size=18)),
                      pad=dict(t=15), ticklen=0,
                      steps=[dict(method='animate', label=f'{t:g}',
                                  args=[[f'{t:g}'], instant])
                             for t in values])])
    return fig

In Chapter 1.4, we drew the vector

\vec v=\begin{bmatrix}3\\-1\end{bmatrix}

along with $3\vec v$ and $-\tfrac12\vec v$ . When drawn from the origin, all three arrows lie on the same line.

# Recreate Chapter 1.4's plot with the larger annotation size used in this section.
from IPython.display import Image, display

fig = style_plotly(go.Figure(), renderer='png')
vdeltay = 0.7
# Draw 3v first so that v remains visible on top of it.
for end, color, label in [
    ((9, -3), 'black', r'$3\vec{v}$'),
    ((3, -1), BLUE, r'$\vec{v}$'),
    ((-1.5, 0.5), 'gray', r'$-\frac{1}{2}\vec{v}$'),
]:
    line2(fig, (0, 0), end, color=color, width=4)
    fig.add_annotation(
        x=end[0], y=end[1], ax=0, ay=0,
        xref='x', yref='y', axref='x', ayref='y',
        showarrow=True, arrowhead=3, arrowsize=1.2, arrowwidth=3,
        arrowcolor=color,
    )
    label2(fig, np.asarray(end) / 2, label, dx=0.4, dy=-vdeltay, color=color)

fig.update_layout(width=800, height=450, yaxis_scaleanchor='x')
fig.update_xaxes(range=[-2, 10], tickvals=np.arange(-2, 11))
fig.update_yaxes(range=[-4, 2], tickvals=np.arange(-4, 3))
fig.layout.annotations[5].x -= 0.25
fig.layout.annotations[5].y += 2 * vdeltay

image_path = _notes_root / 'ch02/imgs/scalar-multiplication-from-1-4.png'
fig.write_image(str(image_path), scale=3)
display(Image(filename=str(image_path)))

What if we drew every scalar multiple of $\vec v$ ? Positive multiples extend along one direction, negative multiples extend along the opposite direction, and $0\vec v$ gives the origin. Together, they fill the entire line!

We’ll call this collection the span of $\vec v$ and connect it to a familiar line equation. Then we’ll use orthogonality from Chapter 1.6 to describe the same line using a perpendicular vector.

Homogeneous and nonhomogeneous equations¶

A line in $\mathbb R^2$ has an equation $ax+by=c$ , where $a$ and $b$ are not both zero. Let’s start with

\ell:\quad 3x-4y=0.

This line passes through the origin: substituting $(0,0)$ gives $3(0)-4(0)=0$ .

fig=base2((-5.5,6),(-4.5,5))
line2(fig,(-5.3,-3.975),(5.7,4.275))
for xy,dx,dy in [((0,0),0.55,-0.45),((4,3),0,0.55),((-4,-3),0.4,-0.55)]:
    point2(fig,xy,str(xy),dx,dy)
label2(fig,(2.1,4.2),r'$3x-4y=0$',color=BLUE)
fig.show(scale=3)

What happens if we change the right-hand side from 0 to 7? The origin no longer satisfies the equation, since $3(0)-4(0)\ne7$ . Instead, $3x-4y=7$ describes a parallel line.

fig=base2((-5.5,7),(-5.5,5.5),width=760)
line2(fig,(-5.2,-3.9),(6.6,4.95))
line2(fig,(-4.8,-5.35),(6.6,3.2),ORANGE)
point2(fig,(0,0),'(0, 0)',dx=-0.7,dy=0.3)
point2(fig,(1,-1),'(1, -1)',dx=0.8,dy=-0.3)
point2(fig,(5,2),'(5, 2)',dx=0.4,dy=-0.5)
label2(fig,(3,4.2),r'$3x-4y=0$',color=BLUE)
label2(fig,(4.9,0.4),r'$3x-4y=7$',color=ORANGE)
fig.show(scale=3)

Keeping $a$ and $b$ fixed while varying $c$ gives parallel lines. For example, $3x-4y=c$ describes every line parallel to $\ell$ .

We’ll first focus on the homogeneous case: lines through the origin.

Lines as spans¶

The opening example suggests a second way to describe a line through the origin: take all scalar multiples of one nonzero vector along it.

For $\ell:3x-4y=0$ , let’s use

\vec v=\begin{bmatrix}4\\3\end{bmatrix}.

Every point on $\ell$ has the form $(4t,3t)$ for some scalar $t\in\mathbb R$ . Equivalently, we can write

\begin{bmatrix}x\\y\end{bmatrix}=t\begin{bmatrix}4\\3\end{bmatrix},\qquad t\in\mathbb R.

This is a parametric form of the line. The scalar $t$ is called the parameter: each value of $t$ gives a point on the line, and letting $t$ range over all real numbers gives the entire line.

Drag the slider below to change $t$ and watch $t\vec v$ sweep along the line. Positive and negative values give opposite directions; when $t=0$ , the vector is the zero vector.

Loading...

Drag the slider: the endpoint $(4t,3t)$ of $t\vec v$ stays on $3x-4y=0$ . The slider shows part of the line; $t$ can be any real number.

So $\ell=\operatorname{span}(\vec v)$ . We could have picked any nonzero vector along this line, such as

\frac12\vec v=\begin{bmatrix}2\\3/2\end{bmatrix},\qquad -\frac12\vec v=\begin{bmatrix}-2\\-3/2\end{bmatrix},\qquad \frac13\vec v=\begin{bmatrix}4/3\\1\end{bmatrix}.

Multiplying by a nonzero scalar changes the length, and possibly reverses the direction, as we saw in Chapter 1.5. It does not change the line that the vector spans:

\ell=\operatorname{span}(\vec v)=\operatorname{span}\left(\frac12\vec v\right) =\operatorname{span}\left(-\frac12\vec v\right)=\operatorname{span}\left(\frac13\vec v\right).

Normal vectors and line equations¶

In Chapter 1.6, we learned that two vectors are orthogonal exactly when their dot product is zero. We can use this fact to connect a line’s equation to its geometry.

Let’s return to $\ell:3x-4y=0$ . We’ll use $\ell^\perp$ to refer to the line through the origin perpendicular to $\ell$ .

fig=base2((-5,6),(-5.5,5.5),height=650)
line2(fig,(-4.8,-3.6),(5.7,4.275),BLUE)
line2(fig,(-3.9,5.2),(3.9,-5.2),ORANGE)
vec2(fig,(4,3),r'$\vec{v}$',BLUE,dx=-0.25,dy=0.65)
vec2(fig,(3,-4),r'$\vec{w}$',ORANGE,dx=0.65,dy=-0.15)
square2(fig,(0,0),(4,3),(3,-4),0.45)
point2(fig,(0,0))
label2(fig,(-3.1,-3.3),r'$\ell$',color=BLUE)
label2(fig,(-2.6,4.3),r'$\ell^\perp$',color=ORANGE)
fig.show(scale=3)

fig=base2((-3.5,4.5),(-4.5,4.5),width=620,height=620)
line2(fig,(-3.3,-1.65),(4.3,2.15),BLUE)
line2(fig,(-2.1,4.2),(2.1,-4.2),ORANGE)
vec2(fig,(2,1),r'$\vec{v}$',BLUE,dx=0,dy=0.6)
vec2(fig,(1,-2),r'$\vec{w}$',ORANGE,dx=0.7,dy=0)
square2(fig,(0,0),(2,1),(1,-2),0.35)
label2(fig,(3.8,1.4),r"$\ell'$",color=BLUE)
label2(fig,(-1.1,3.4),r"$(\ell')^\perp$",color=ORANGE)
fig.show(scale=3)

The coefficients of $3x-4y=0$ give us the vector

\vec w=\begin{bmatrix}3\\-4\end{bmatrix}.

Why does this vector lie on $\ell^\perp$ ? We can rewrite the equation of $\ell$ as a dot product:

\begin{bmatrix}3\\-4\end{bmatrix}\cdot\begin{bmatrix}x\\y\end{bmatrix}=0.

So $(x,y)$ lies on $\ell$ exactly when $\begin{bmatrix}x\\y\end{bmatrix}$ is orthogonal to $\vec w$ .

We now have two ways to describe $\ell$ : take multiples of a vector along it, or use a vector perpendicular to it. With $\vec v=\begin{bmatrix}4\\3\end{bmatrix}$ and $\vec w=\begin{bmatrix}3\\-4\end{bmatrix}$ ,

\ell=\operatorname{span}(\vec v) =\left\{\begin{bmatrix}x\\y\end{bmatrix}:\vec w\cdot\begin{bmatrix}x\\y\end{bmatrix}=0\right\}.

The roles reverse for $\ell^\perp$ : $\vec w$ lies along it, and $\vec v$ is perpendicular to it. Therefore,

\ell^\perp=\operatorname{span}(\vec w) =\left\{\begin{bmatrix}x\\y\end{bmatrix}:\vec v\cdot\begin{bmatrix}x\\y\end{bmatrix}=0\right\}.

Taking the dot product with $\vec v$ gives the equation $4x+3y=0$ for $\ell^\perp$ .

A normal vector is not unique. For example, $3\vec w=\begin{bmatrix}9\\-12\end{bmatrix}$ is also normal to $\ell$ , giving $9x-12y=0$ . This is just the original equation multiplied by 3, so it describes the same line.

Activity 2

Consider the line

\ell'=\operatorname{span}\left(\begin{bmatrix}2\\1\end{bmatrix}\right).

Draw $\ell'$ and $(\ell')^\perp$ .
Find a nonzero vector on $(\ell')^\perp$ .
Use your vector from the previous subpart to write an equation for $\ell'$ .
Find another nonzero vector on $(\ell')^\perp$ .
Use your vector from the previous subpart to write another equation for $\ell'$ .
How do your equations for $\ell'$ differ?
Write down an equation for $(\ell')^\perp$ .

Solution

The line $\ell'$ has slope $1/2$ , and $(\ell')^\perp$ has slope -2. Both pass through the origin, as shown below.
One choice is $\vec w=\begin{bmatrix}1\\-2\end{bmatrix}$ , since $\begin{bmatrix}2\\1\end{bmatrix}\cdot\begin{bmatrix}1\\-2\end{bmatrix}=2-2=0$ .
Using $\vec w$ as a normal vector gives $x-2y=0$ .
Another choice is $2\vec w=\begin{bmatrix}2\\-4\end{bmatrix}$ .
This gives $2x-4y=0$ .
The second equation is twice the first, so they have the same solution set.
The vector $\begin{bmatrix}2\\1\end{bmatrix}$ is normal to $(\ell')^\perp$ , giving $2x+y=0$ .

Affine lines¶

Now let’s return to $3x-4y=7$ , which does not pass through the origin.

In Chapter 1.4, we saw that a vector represents a displacement and can be drawn from any starting point. We can use that idea to shift an entire line.

fig=base2((-3,7),(-4,5.5),width=720)
line2(fig,(-2.8,-2.1),(6.6,4.95),BLUE)
line2(fig,(-2.8,-3.85),(6.6,3.2),ORANGE)
vec2(fig,(5,2),r'$\vec{v}$',ORANGE,start=(1,-1),dx=0.1,dy=0.65)
point2(fig,(1,-1),'(1, -1)',dx=0,dy=-0.6)
point2(fig,(5,2),'(5, 2)',dx=0.8,dy=-0.35)
label2(fig,(2.1,3.7),r'$3x-4y=0$',color=BLUE)
label2(fig,(4.2,-0.8),r'$3x-4y=7$',color=ORANGE)
fig.show(scale=3)

The point $(1,-1)$ lies on $3x-4y=7$ . Starting there and adding any multiple of $\vec v=\begin{bmatrix}4\\3\end{bmatrix}$ takes us to another point on the line. The vector $\vec v$ gives its direction, just as it did for $3x-4y=0$ .

So we can describe the line as

\begin{bmatrix}1\\-1\end{bmatrix}+\operatorname{span}(\vec v) =\left\{\begin{bmatrix}1\\-1\end{bmatrix}+t\vec v:t\in\mathbb R\right\}.

This gives the parametric form

\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}+t\begin{bmatrix}4\\3\end{bmatrix},\qquad t\in\mathbb R.

Our starting point is not unique. For example, $(5,2)$ also lies on the line, so the same line is

\left\{\begin{bmatrix}5\\2\end{bmatrix}+t\vec v:t\in\mathbb R\right\}.

Finding the constant from a point¶

Suppose we know a point on an affine line and a normal vector, but not its equation. How can we find the constant on the right-hand side?

Let $\vec p$ be the position vector of the known point, let $\vec n$ be a nonzero normal vector, and write

\vec x=\begin{bmatrix}x\\y\end{bmatrix}.

A point with position vector $\vec x$ lies on the line exactly when the displacement $\vec x-\vec p$ is perpendicular to $\vec n$ . Therefore,

\vec n\cdot(\vec x-\vec p)=0 \quad\Longleftrightarrow\quad \vec n\cdot\vec x=\vec n\cdot\vec p.

In general, an affine line in $\mathbb R^2$ through $\vec p$ with nonzero normal vector $\vec n$ has equation

\vec n\cdot\vec x=\vec n\cdot\vec p.

The right-hand side is a constant: take the dot product of the normal vector with the known point’s position vector. For our line, $\vec p=\begin{bmatrix}1\\-1\end{bmatrix}$ and $\vec n=\begin{bmatrix}3\\-4\end{bmatrix}$ give $3(1)-4(-1)=7$ , recovering $3x-4y=7$ .

For another example, consider the affine line in $\mathbb R^2$

\ell=\vec p+\operatorname{span}(\vec v),\qquad \vec p=\begin{bmatrix}2\\-1\end{bmatrix},\qquad \vec v=\begin{bmatrix}1\\2\end{bmatrix}.

To write an equation for this line, choose a nonzero vector perpendicular to its direction:

\vec n=\begin{bmatrix}2\\-1\end{bmatrix},\qquad \vec n\cdot\vec v=2(1)-1(2)=0.

Then use the known point to find the constant:

\vec n\cdot\vec p=2(2)-1(-1)=5.

The dot-product equation and its expanded form are

\begin{bmatrix}2\\-1\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix}=5, \qquad 2x-y=5.

To check the result, substitute the parametric coordinates of the line:

2(2+t)-(-1+2t)=5.

This holds for every $t\in\mathbb R$ , as it should.

fig=base2((-1,6),(-4,6),width=600,height=660,tick=1)
line2(fig,(0.5,-4),(5.5,6),BLUE)
point2(fig,(2,-1),'(2, -1)',dx=-0.8,dy=-0.45)
vec2(fig,(3,1),r'$\vec v$',BLUE,start=(2,-1),dx=-0.5,dy=0.15)
vec2(fig,(4,-2),r'$\vec n$',ORANGE,start=(2,-1),dx=0.15,dy=0.55)
square2(fig,(2,-1),(1,2),(2,-1),0.35)
label2(fig,(4.5,4),r'$2x-y=5$',dx=-1.1,dy=0.5,color=BLUE)
fig.show(scale=2)

Here’s a video reviewing how to find the equation of a line in $\mathbb{R}^2$ .