Fluency exercise

## Solution

Here are the equations of 12 straight lines.

$\mathbf{r}= \begin{pmatrix} 0 \\ -6 \\ \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 4 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ -3 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 8 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 2 \\ 9.5 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -8 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 6 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 6 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 10 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} -4 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} 5 \\ -1 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 6 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 11 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -6 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \end{pmatrix}$

These 12 straight lines can be divided up into six pairs, each pair matching one of the following descriptions:

• These lines are parallel.

• These lines are perpendicular.

• These lines have the same $y$-intercept.

• These lines have the same $x$-intercept.

• These lines both go through the point $(1, 5)$.

• These lines …

Can you sort them into the correct pairs and complete the final description?

While sketching the lines could be helpful, 12 might be too many on one graph. Are there any pairs we can match up by just looking at the equations?

$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 6 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 6 \end{pmatrix}$

Equations 7 and 10 have the same position vector which also happens to be their $y$-interept.

Do you think there are particular features of lines that are easier to spot when the equation is written in vector form compared to Cartesian form?

$\mathbf{r}= \begin{pmatrix} 2 \\ 9.5 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -8 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 4 \end{pmatrix}$

Equations 3 and 6 are parallel lines, but it looks like they have different direction vectors.

How do you know these lines are parallel? What is the same about the two direction vectors?

$\mathbf{r}= \begin{pmatrix} 0 \\ -6 \\ \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 4 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \end{pmatrix}$

Equations 1 and 5 are perpendicular lines.

Can you use a diagram to convince yourself that the two are perpendicular?

Can you find a different direction vector for equation 1 that would still make it perpendicular to equation 5?

Now we’ve paired up 6 equations, a sketch of the remaining 6 will be easier to work with.

Depending on how accurate our sketch is, we may be able to write down some of the pairs straight away, or we might need to check the intersection points. For example, it looks like equations 4 and 12 share the same $x$-intercept.

$\mathbf{r}= \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \end{pmatrix}$

We could check by finding the value of $\mu$ for each equation that gives $y = 0$, or we could solve the equations simultaneously.

Writing them as simultaneous equations gives us $-2\mu = 1+3\lambda$ and $2+\mu = -2+2\lambda$, which solve to give $\mu = -2$ and $\lambda=1$. Substituting either of these values into their equation confirms that the two lines meet at their $x$-intercept of $(4,0)$.

$\mu$ was changed into $\lambda$ for the second equation. Why? What would happen if we used the same variable for both equations?

From the graph it looks like equation 2 and 11 meet at the point $(1, 5).$

$\mathbf{r}= \begin{pmatrix} 0 \\ -3 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 8 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} 0 \\ 11 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -6 \end{pmatrix}$

We could solve the equations for their intersection points, or we could notice that setting $\mu = 1$ for both equations gives an $x$-coordinate of $1$. Does it give the correct $y$-coordinate for both lines?

$\mathbf{r}= \begin{pmatrix} 0 \\ 6 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 10 \end{pmatrix}$

$\mathbf{r}= \begin{pmatrix} -4 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} 5 \\ -1 \end{pmatrix}$

What do the final two lines have in common?

#### Taking it further

Can you come up with the equation of a line that would match with 2 different pairs at the same time?

If the lines were given in three dimensions would you still be able to match up pairs with the same properties? Which properties might be the most challenging to match; parallel, perpendicular, shared $x$-intercepts, shared $y$-intercepts or a point of intersection?