\(k \parallel l \text{,}\) dus \(l{:}\,-6 x - 5 y = c \text{.}\)

\(\begin{rcases}-6 x - 5 y = c \\ \text{door } A (3 , 7)\end{rcases} c = -6 ⋅ 3 - 5 ⋅ 7 = -53\)
Dus \(l{:}\,-6 x - 5 y = -53 \text{.}\)

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

Aftrekken geeft \(-x = -4\) dus \(x = 4 \text{.}\)

\(\begin{rcases}5 x + 4 y = 2 \\ x = 4\end{rcases} \begin{matrix}5 ⋅ 4 + 4 y = 2 \\ 4 y = -18 \\ y = -4\frac{1}{2}\end{matrix}\)

Dus \(S (4 , -4\frac{1}{2}) \text{.}\)

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

\(3 x + 3 x + 1 = -3\)
\(6 x = -4\)
Dus \(x = -\frac{2}{3} \text{.}\)

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

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

\(k{:}\,4 x - 7 y = -8\) omschrijven geeft \(y = \frac{4}{7} x + 1\frac{1}{7}\) dus \(\text{rc}_{k} = \frac{4}{7} \text{.}\)
\(l{:}\,x + 6 y = -3\) omschrijven geeft \(y = -\frac{1}{6} x - \frac{1}{2}\) dus \(\text{rc}_{l} = -\frac{1}{6} \text{.}\)

\(\tan(\alpha ) = \frac{4}{7}\) geeft \(\alpha = \tan^{-1}(\frac{4}{7}) = 29{,}74...\degree \text{.}\)
\(\tan(\beta ) = -\frac{1}{6}\) geeft \(\beta = \tan^{-1}(-\frac{1}{6}) = -9{,}46...\degree \text{.}\)

\(\varphi = \alpha - \beta = 29{,}74...\degree - -9{,}46...\degree = 39{,}20...\degree \text{,}\) dus de gevraagde hoek is \(39{,}2\degree \text{.}\)

\(\frac{1}{2} ≠ \frac{3}{4} ≠ \frac{2}{5} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({6 \over 18} = {p \over -6} = {5 \over q}\)

\({6 \over 18} = {p \over -6}\) geeft \(p = -2\) (en \({6 \over 18} = {5 \over q}\) geeft \(q = 15 \text{)}\)

Evenwijdig, dus \(p = -2\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,2 x + 3 y = c \text{.}\)

\(\begin{rcases}2 x + 3 y = c \\ \text{door } A (4 , -8)\end{rcases} c = 2 ⋅ 4 + 3 ⋅ -8 = -16\)
Dus \(l{:}\,2 x + 3 y = -16 \text{.}\)