\(k \parallel l \text{,}\) dus \(l{:}\,-8 x - 6 y = c \text{.}\)

\(\begin{rcases}-8 x - 6 y = c \\ \text{door } A (3 , 9)\end{rcases} c = -8 ⋅ 3 - 6 ⋅ 9 = -78\)
Dus \(l{:}\,-8 x - 6 y = -78 \text{.}\)

\(\begin{cases}5 x - 4 y = 3 \\ 3 x - 2 y = 2\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}5 x - 4 y = 3 \\ 6 x - 4 y = 4\end{cases}\)

Aftrekken geeft \(-x = -1\) dus \(x = 1 \text{.}\)

\(\begin{rcases}5 x - 4 y = 3 \\ x = 1\end{rcases} \begin{matrix}5 ⋅ 1 - 4 y = 3 \\ -4 y = -2 \\ y = \frac{1}{2}\end{matrix}\)

Dus \(S (1 , \frac{1}{2}) \text{.}\)

Substitutie geeft \(2 x - 2 (2 x + 2) = 3\)

\(2 x - 4 x - 4 = 3\)
\(-2 x = 7\)
Dus \(x = -3\frac{1}{2} \text{.}\)

\(\begin{rcases}y = 2 x + 2 \\ x = -3\frac{1}{2}\end{rcases} y = 2 ⋅ -3\frac{1}{2} + 2 = -5\)

Dus \(S (-3\frac{1}{2} , -5) \text{.}\)

\(k{:}\,2 x + 5 y = 1\) omschrijven geeft \(y = -\frac{2}{5} x + \frac{1}{5}\) dus \(\text{rc}_{k} = -\frac{2}{5} \text{.}\)
\(l{:}\,9 x + 7 y = 8\) omschrijven geeft \(y = -1\frac{2}{7} x + 1\frac{1}{7}\) dus \(\text{rc}_{l} = -1\frac{2}{7} \text{.}\)

\(\tan(\alpha ) = -\frac{2}{5}\) geeft \(\alpha = \tan^{-1}(-\frac{2}{5}) = -21{,}80...\degree \text{.}\)
\(\tan(\beta ) = -1\frac{2}{7}\) geeft \(\beta = \tan^{-1}(-1\frac{2}{7}) = -52{,}12...\degree \text{.}\)

\(\varphi = \alpha - \beta = -21{,}80...\degree - -52{,}12...\degree = 30{,}32...\degree \text{,}\) dus de gevraagde hoek is \(30{,}3\degree \text{.}\)

\(\frac{5}{15} = \frac{6}{18} = \frac{4}{12} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

\({4 \over 8} = {p \over -10} = {q \over -6}\)

\({4 \over 8} = {p \over -10}\) geeft \(p = -5\) en \({4 \over 8} = {q \over -6}\) geeft \(q = -3\)

Samenvallen, dus \(p = -5\) en \(q = -3 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,-4 x + 7 y = c \text{.}\)

\(\begin{rcases}-4 x + 7 y = c \\ \text{door } A (-2 , 1)\end{rcases} c = -4 ⋅ -2 + 7 ⋅ 1 = 15\)
Dus \(l{:}\,-4 x + 7 y = 15 \text{.}\)