\(k \parallel l \text{,}\) dus \(l{:}\,-4 x - 2 y = c \text{.}\)

\(\begin{rcases}-4 x - 2 y = c \\ \text{door } A (8 , 7)\end{rcases} c = -4 ⋅ 8 - 2 ⋅ 7 = -46\)
Dus \(l{:}\,-4 x - 2 y = -46 \text{.}\)

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

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

\(\begin{rcases}3 x + 2 y = 1 \\ x = -2\end{rcases} \begin{matrix}3 ⋅ -2 + 2 y = 1 \\ 2 y = 7 \\ y = 3\frac{1}{2}\end{matrix}\)

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

Substitutie geeft \(3 x - 3 (3 x - 1) = 5\)

\(3 x - 9 x + 3 = 5\)
\(-6 x = 2\)
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 \(S (-\frac{1}{3} , -2) \text{.}\)

\(k{:}\,9 x + 4 y = 3\) omschrijven geeft \(y = -2\frac{1}{4} x + \frac{3}{4}\) dus \(\text{rc}_{k} = -2\frac{1}{4} \text{.}\)
\(l{:}\,x + 7 y = -5\) omschrijven geeft \(y = -\frac{1}{7} x - \frac{5}{7}\) dus \(\text{rc}_{l} = -\frac{1}{7} \text{.}\)

\(\tan(\alpha ) = -2\frac{1}{4}\) geeft \(\alpha = \tan^{-1}(-2\frac{1}{4}) = -66{,}03...\degree \text{.}\)
\(\tan(\beta ) = -\frac{1}{7}\) geeft \(\beta = \tan^{-1}(-\frac{1}{7}) = -8{,}13...\degree \text{.}\)

\(\varphi = \alpha - \beta = -66{,}03...\degree - -8{,}13...\degree = -57{,}90...\degree \text{,}\) dus de gevraagde hoek is \(57{,}9\degree \text{.}\)

\(-\frac{1}{2} = -\frac{5}{10} ≠ \frac{3}{2} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

\({1 \over \frac{1}{3}} = {p \over -\frac{5}{3}} = {q \over 2}\)

\({1 \over \frac{1}{3}} = {p \over -\frac{5}{3}}\) geeft \(p = -5\) (en \({1 \over \frac{1}{3}} = {q \over 2}\) geeft \(q = 6 \text{)}\)

Evenwijdig, dus \(p = -5\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,-5 x - 9 y = c \text{.}\)

\(\begin{rcases}-5 x - 9 y = c \\ \text{door } A (-4 , 3)\end{rcases} c = -5 ⋅ -4 - 9 ⋅ 3 = -7\)
Dus \(l{:}\,-5 x - 9 y = -7 \text{.}\)