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

\(\begin{rcases}-3 x - 2 y = c \\ \text{door } A (-4 , 9)\end{rcases} c = -3 ⋅ -4 - 2 ⋅ 9 = -6\)
Dus \(l{:}\,-3 x - 2 y = -6 \text{.}\)

\(\begin{cases}4 x - 2 y = 1 \\ 5 x - 4 y = -4\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}8 x - 4 y = 2 \\ 5 x - 4 y = -4\end{cases}\)

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

\(\begin{rcases}4 x - 2 y = 1 \\ x = 2\end{rcases} \begin{matrix}4 ⋅ 2 - 2 y = 1 \\ -2 y = -7 \\ y = 3\frac{1}{2}\end{matrix}\)

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

Substitutie geeft \(2 x - 2 (-2 x + 1) = -5\)

\(2 x + 4 x - 2 = -5\)
\(6 x = -3\)
Dus \(x = -\frac{1}{2} \text{.}\)

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

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

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

\(\tan(\alpha ) = \frac{3}{7}\) geeft \(\alpha = \tan^{-1}(\frac{3}{7}) = 23{,}19...\degree \text{.}\)
\(\tan(\beta ) = \frac{1}{9}\) geeft \(\beta = \tan^{-1}(\frac{1}{9}) = 6{,}34...\degree \text{.}\)

\(\varphi = \alpha - \beta = 23{,}19...\degree - 6{,}34...\degree = 16{,}85...\degree \text{,}\) dus de gevraagde hoek is \(16{,}9\degree \text{.}\)

\(\frac{5}{10} ≠ \frac{2}{6} ≠ -\frac{1}{3} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

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

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

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

\(k \perp l \text{,}\) dus \(l{:}\,-9 x + 2 y = c \text{.}\)

\(\begin{rcases}-9 x + 2 y = c \\ \text{door } A (-5 , -4)\end{rcases} c = -9 ⋅ -5 + 2 ⋅ -4 = 37\)
Dus \(l{:}\,-9 x + 2 y = 37 \text{.}\)