[Richtingscoëfficiënt is de stijging per stap naar rechts, dus]
\(\overrightarrow{r} = \begin{pmatrix}1 \\ 7\end{pmatrix} \text{.}\)

\(\overrightarrow{s} = \begin{pmatrix}0 \\ 2\end{pmatrix} \text{,}\) dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 2\end{pmatrix} + t ⋅ \begin{pmatrix}1 \\ 7\end{pmatrix} \text{.}\)

\(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-6 \\ 0\end{pmatrix} + t ⋅ \begin{pmatrix}-3 \\ -2\end{pmatrix} \text{.}\)

\(l \text{: } x = 2 - 5 t ∧ y = 4 - 6 t \text{.}\)

\(\begin{cases}x = -5 t + 1 \\ y = -6 t - 4\end{cases}\) \(\begin{vmatrix}6 \\ 5\end{vmatrix}\) geeft \(\begin{cases}6 x = -30 t + 6 \\ 5 y = -30 t - 20\end{cases}\)

Aftrekken geeft
\(l{:}\,6 x - 5 y = 26 \text{.}\)

\(l \text{: } x = t ∧ y = -3 t - 1 \text{.}\)

\(x = 5 t + 1\) geeft
\(y = -2 ⋅ (5 t + 1) + 6\)
\(\text{} = -10 t - 2 + 6\)
\(\text{} = -10 t + 4 \text{.}\)

\(l \text{: } x = 5 t + 1 ∧ y = -10 t + 4 \text{.}\)

\(\overrightarrow{r}_{l} = \begin{pmatrix}-3 \\ 5\end{pmatrix} \text{,}\) dus \(\overrightarrow{n}_{l} = \begin{pmatrix}5 \\ 3\end{pmatrix} \text{.}\)

\(\begin{rcases}5 x + 3 y = c \\ \text{door } (7 , 0)\end{rcases} \begin{matrix}c = 5 ⋅ 7 + 3 ⋅ 0\end{matrix} 35\)

Dus \(5 x + 3 y = 35 \text{.}\)

\(\overrightarrow{n}_{l} = \begin{pmatrix}4 \\ -5\end{pmatrix} \text{,}\) dus \(\overrightarrow{r}_{l} = \begin{pmatrix}5 \\ 4\end{pmatrix} \text{.}\)

\(\begin{rcases}l{:}\,4 x - 5 y = -3 \\ x = 0\end{rcases} \begin{matrix}4 ⋅ 0 - 5 y = -3 \\ y = \frac{3}{5}\end{matrix}\)

Dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ \frac{3}{5}\end{pmatrix} + t ⋅ \begin{pmatrix}5 \\ 4\end{pmatrix} \text{.}\)

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

\(\overrightarrow{s} = \begin{pmatrix}0 \\ -2\end{pmatrix} \text{,}\) dus \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ -2\end{pmatrix} + t ⋅ \begin{pmatrix}6 \\ -5\end{pmatrix} \text{.}\)

\(\overrightarrow{r}_{k} = \begin{pmatrix}-2 \\ -3\end{pmatrix} \text{,}\) dus \(\overrightarrow{n}_{k} = \begin{pmatrix}3 \\ -2\end{pmatrix} \text{.}\)

\(\begin{rcases}3 x - 2 y = c \\ \text{door } A (0 , 1)\end{rcases} \begin{matrix}c = 3 ⋅ 0 - 2 ⋅ 1 = -2\end{matrix}\)

Dus \(3 x - 2 y = -2 \text{.}\)

\(\overrightarrow{r}_{k} = \begin{pmatrix}4 \\ 5\end{pmatrix} \text{,}\) dus \(4 x + 5 y = c\) staat loodrecht.

\(\begin{rcases}4 x + 5 y = c \\ \text{door } A (0 , -7)\end{rcases} \begin{matrix}c = 4 ⋅ 0 + 5 ⋅ -7 = -35\end{matrix}\)

Dus \(4 x + 5 y = -35 \text{.}\)

\(\overrightarrow{r}_{l} = \overrightarrow{n}_{k} = \begin{pmatrix}5 \\ 3\end{pmatrix} \text{.}\)

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