We’ve done a fair amount of landscaping, as it were—running out a long way in the field of linear algebra to mark interesting points. Now it seems like a good time to turn back and start tending to each area a little more closely. An excellent way to do that—to make things more secure—is to practice.

The sets below can be completed after reading Lines the Linear Algebra Way. I would suggest completing this by working on a piece of paper and then checking your answers—one at a time at first and then go a stretch before checking. Try to complete each exercise before either checking back with the original post or looking at my answer. My answers can be uncovered by hovering under each exercise.

Convert each equation of a line to a vector equation in parametric form, \(\mathtt{l(k)=p+kv}\), where \(\mathtt{p}\) and \(\mathtt{v}\) are 2D vectors and \(\mathtt{k}\) is a scalar variable.

- \(\mathtt{y=x+3}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{y=2x+3}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{y=-2x}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-2}\end{bmatrix}\mathtt{k}\), or just \(\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-2}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{y=x}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}\), or just \(\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{y=10}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{10}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{0}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{x+y=1}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{1}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-1}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{2x+3y=9}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,3}\\\mathtt{-2}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{x=5}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{5}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{1}\end{bmatrix}\mathtt{k}\)

- \(\mathtt{-3x-\frac{1}{4}y=15}\)

\(\mathtt{l(k)=}\begin{bmatrix}\mathtt{\,\,\,\,0}\\\mathtt{-60}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-12}\end{bmatrix}\mathtt{k}\)

- Devise an algorithm you could use to convert an equation for a line in slope-intercept form to a vector equation for the line in parametric form.

For slope-intercept form, \(\mathtt{y=mx+b}\), the conversion \(\begin{bmatrix}\mathtt{0}\\\mathtt{b}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{m}\end{bmatrix}\mathtt{k}\,\) seems to work for most lines.

There are many ways to write the above vector equations. In particular, the intercept vector does not have to have a first component of \(\mathtt{0}\), and the slope vector does not have to be in simplest form. All that is required is for the intercept vector to get you to some point on the line and for the slope vector to correctly represent the slope of the line.

Now let’s go the other way. Identify the slope, y-intercept, and x-intercept of each line.

- \(\mathtt{l_{1}(k)=}\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,2}\\\mathtt{-1}\end{bmatrix}\mathtt{k}\)

slope → \(\mathtt{-\frac{1}{2}}\), y-intercept → \(\mathtt{\frac{5}{2}}\), x-intercept → \(\mathtt{5}\)

- \(\mathtt{l_{2}(k)=}\begin{bmatrix}\mathtt{-3}\\\mathtt{-5}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{4}\end{bmatrix}\mathtt{k}\)

slope → undefined, y-intercept → none, x-intercept → \(\mathtt{-3}\)

- \(\mathtt{l_{3}(k)=}\begin{bmatrix}\mathtt{\,\,\,\,0}\\\mathtt{-1}\end{bmatrix}+\begin{bmatrix}\mathtt{2}\\\mathtt{2}\end{bmatrix}\mathtt{k}\)

slope → \(\mathtt{1}\), y-intercept → \(\mathtt{-1}\), x-intercept → \(\mathtt{1}\)

- \(\mathtt{l_{4}(k)=}\begin{bmatrix}\mathtt{\,\,\,\,3}\\\mathtt{-9}\end{bmatrix}+\begin{bmatrix}\mathtt{7}\\\mathtt{0}\end{bmatrix}\mathtt{k}\)

slope → \(\mathtt{0}\), y-intercept → \(\mathtt{-9}\), x-intercept → none

- \(\mathtt{l_{5}(k)=}\begin{bmatrix}\mathtt{-5}\\\mathtt{\,\,\,\,1}\end{bmatrix}+\begin{bmatrix}\mathtt{-3}\\\mathtt{\,\,\,\,4}\end{bmatrix}\mathtt{k}\)

slope → \(\mathtt{-\frac{4}{3}}\), y-intercept → \(\mathtt{-\frac{17}{3}}\), x-intercept → \(\mathtt{-\frac{17}{4}}\)

And just two more. Determine if each point is on the line mentioned from above.

- Is \(\mathtt{(-11,3)}\) on the line \(\mathtt{l_1}\)?

No: \(\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,2}\\\mathtt{-1}\end{bmatrix}\mathtt{k=}\begin{bmatrix}\mathtt{-11}\\\mathtt{\,\,\,\,3}\end{bmatrix}\).

No value of \(\mathtt{k}\) solves both \(\mathtt{2k+1=-11}\) and \(\mathtt{-k+2=3}\).

- Is \(\mathtt{(5,5)}\) on the line \(\mathtt{l_3}\)?

No. Since the line has a slope of \(\mathtt{1}\), it would contain \(\mathtt{(5,5)}\) if the line passed through the origin, but it does not.