\(k \parallel l \text{,}\) dus \(l{:}\,-7 x + 6 y = c \text{.}\)

\(\begin{rcases}-7 x + 6 y = c \\ \text{door } A (2 , 3)\end{rcases} c = -7 ⋅ 2 + 6 ⋅ 3 = 4\)
Dus \(l{:}\,-7 x + 6 y = 4 \text{.}\)

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

\(4 x + 6 x - 6 = -1\)
\(10 x = 5\)
Dus \(x = \frac{1}{2} \text{.}\)

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

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

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

\(\tan(\alpha ) = 2\) geeft \(\alpha = \tan^{-1}(2) = 63{,}43...\degree \text{.}\)
\(\tan(\beta ) = \frac{1}{8}\) geeft \(\beta = \tan^{-1}(\frac{1}{8}) = 7{,}12...\degree \text{.}\)

\(\varphi = \alpha - \beta = 63{,}43...\degree - 7{,}12...\degree = 56{,}30...\degree \text{,}\) dus de gevraagde hoek is \(56{,}3\degree \text{.}\)

\(\frac{1}{3} = \frac{4}{12} = \frac{3}{9} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

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

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

Evenwijdig, dus \(p = \frac{2}{3}\) en \(q\) mag elk getal zijn.

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

\(\begin{rcases}-9 x - y = c \\ \text{door } A (4 , 3)\end{rcases} c = -9 ⋅ 4 - 1 ⋅ 3 = -39\)
Dus \(l{:}\,-9 x - y = -39 \text{.}\)