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

\(\begin{rcases}x + 6 y = c \\ \text{door } A (-5 , 9)\end{rcases} c = 1 ⋅ -5 + 6 ⋅ 9 = 49\)
Dus \(l{:}\,x + 6 y = 49 \text{.}\)

\(\begin{cases}5 x + 4 y = 0 \\ x + 2 y = 3\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}5 x + 4 y = 0 \\ 2 x + 4 y = 6\end{cases}\)

Aftrekken geeft \(3 x = -6\) dus \(x = -2 \text{.}\)

\(\begin{rcases}5 x + 4 y = 0 \\ x = -2\end{rcases} \begin{matrix}5 ⋅ -2 + 4 y = 0 \\ 4 y = 10 \\ y = 2\frac{1}{2}\end{matrix}\)

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

Substitutie geeft \(3 x - 2 (3 x - 1) = 3\)

\(3 x - 6 x + 2 = 3\)
\(-3 x = 1\)
Dus \(x = -\frac{1}{3} \text{.}\)

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

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

\(k{:}\,2 x + 4 y = -9\) omschrijven geeft \(y = -\frac{1}{2} x - 2\frac{1}{4}\) dus \(\text{rc}_{k} = -\frac{1}{2} \text{.}\)
\(l{:}\,7 x + y = -8\) omschrijven geeft \(y = -7 x - 8\) dus \(\text{rc}_{l} = -7 \text{.}\)

\(\tan(\alpha ) = -\frac{1}{2}\) geeft \(\alpha = \tan^{-1}(-\frac{1}{2}) = -26{,}56...\degree \text{.}\)
\(\tan(\beta ) = -7\) geeft \(\beta = \tan^{-1}(-7) = -81{,}86...\degree \text{.}\)

\(\varphi = \alpha - \beta = -26{,}56...\degree - -81{,}86...\degree = 55{,}30...\degree \text{,}\) dus de gevraagde hoek is \(55{,}3\degree \text{.}\)

\(-\frac{2}{6} = -\frac{5}{15} ≠ -\frac{3}{4} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

\({1 \over \frac{1}{4}} = {2 \over p} = {q \over -\frac{3}{4}}\)

\({1 \over \frac{1}{4}} = {2 \over p}\) geeft \(p = \frac{1}{2}\) en \({1 \over \frac{1}{4}} = {q \over -\frac{3}{4}}\) geeft \(q = -3\)

Geen punt gemeenschappelijk, dus \(p = \frac{1}{2}\) en \(q ≠ -3 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,-7 x - 5 y = c \text{.}\)

\(\begin{rcases}-7 x - 5 y = c \\ \text{door } A (8 , -4)\end{rcases} c = -7 ⋅ 8 - 5 ⋅ -4 = -36\)
Dus \(l{:}\,-7 x - 5 y = -36 \text{.}\)