[Richtingscoëfficiënt is de stijging per stap naar rechts, dus]
\(\overrightarrow{r} = \begin{pmatrix}1 \\ 6\end{pmatrix} \text{.}\)

\(\overrightarrow{s} = \begin{pmatrix}0 \\ 3\end{pmatrix} \text{,}\) dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 3\end{pmatrix} + t ⋅ \begin{pmatrix}1 \\ 6\end{pmatrix} \text{.}\)

\(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-4 \\ -5\end{pmatrix} + t ⋅ \begin{pmatrix}-3 \\ 7\end{pmatrix} \text{.}\)

\(l \text{: } x = -6 t - 2 ∧ y = 3 t \text{.}\)

\(\begin{cases}x = 7 t - 4 \\ y = -3 t + 6\end{cases}\) \(\begin{vmatrix}3 \\ 7\end{vmatrix}\) geeft \(\begin{cases}3 x = 21 t - 12 \\ 7 y = -21 t + 42\end{cases}\)

Optellen geeft
\(l{:}\,3 x + 7 y = 30 \text{.}\)

\(l \text{: } x = t ∧ y = 2 - 4 t \text{.}\)

\(x = 6 t - 4\) geeft
\(y = 7 ⋅ (6 t - 4) - 5\)
\(\text{} = 42 t - 28 - 5\)
\(\text{} = 42 t - 33 \text{.}\)

\(l \text{: } x = 6 t - 4 ∧ y = 42 t - 33 \text{.}\)

\(\overrightarrow{r}_{l} = \begin{pmatrix}6 \\ -3\end{pmatrix} \text{,}\) dus \(\overrightarrow{n}_{l} = \begin{pmatrix}3 \\ 6\end{pmatrix} \text{.}\)

\(\begin{rcases}3 x + 6 y = c \\ \text{door } (-2 , -7)\end{rcases} \begin{matrix}c = 3 ⋅ -2 + 6 ⋅ -7\end{matrix} -48\)

Dus \(3 x + 6 y = -48 \text{.}\)

\(\overrightarrow{n}_{l} = \begin{pmatrix}5 \\ 6\end{pmatrix} \text{,}\) dus \(\overrightarrow{r}_{l} = \begin{pmatrix}6 \\ -5\end{pmatrix} \text{.}\)

\(\begin{rcases}l{:}\,5 x + 6 y = 3 \\ x = 0\end{rcases} \begin{matrix}5 ⋅ 0 + 6 y = 3 \\ y = \frac{1}{2}\end{matrix}\)

Dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ \frac{1}{2}\end{pmatrix} + t ⋅ \begin{pmatrix}6 \\ -5\end{pmatrix} \text{.}\)

\(\overrightarrow{n}_{k} = \begin{pmatrix}3 \\ -4\end{pmatrix} \text{,}\) dus \(\overrightarrow{r}_{k} = \begin{pmatrix}4 \\ 3\end{pmatrix} \text{.}\)

\(\overrightarrow{s} = \begin{pmatrix}6 \\ -5\end{pmatrix} \text{,}\) dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}6 \\ -5\end{pmatrix} + t ⋅ \begin{pmatrix}4 \\ 3\end{pmatrix} \text{.}\)

\(\overrightarrow{r}_{k} = \begin{pmatrix}2 \\ 6\end{pmatrix} \text{,}\) dus \(\overrightarrow{n}_{k} = \begin{pmatrix}6 \\ -2\end{pmatrix} \text{.}\)

\(\begin{rcases}6 x - 2 y = c \\ \text{door } A (0 , -3)\end{rcases} \begin{matrix}c = 6 ⋅ 0 - 2 ⋅ -3 = 6\end{matrix}\)

Dus \(6 x - 2 y = 6 \text{.}\)

\(\overrightarrow{r}_{k} = \begin{pmatrix}6 \\ -2\end{pmatrix} \text{,}\) dus \(6 x - 2 y = c\) staat loodrecht.

\(\begin{rcases}6 x - 2 y = c \\ \text{door } A (3 , -5)\end{rcases} \begin{matrix}c = 6 ⋅ 3 - 2 ⋅ -5 = 28\end{matrix}\)

Dus \(6 x - 2 y = 28 \text{.}\)

\(\overrightarrow{r}_{l} = \overrightarrow{n}_{k} = \begin{pmatrix}1 \\ -7\end{pmatrix} \text{.}\)

\(\overrightarrow{s} = \begin{pmatrix}6 \\ -4\end{pmatrix} \text{,}\) dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}6 \\ -4\end{pmatrix} + t ⋅ \begin{pmatrix}1 \\ -7\end{pmatrix} \text{.}\)