\(k \parallel l \text{,}\) dus \(l{:}\,-9 x + 7 y = c \text{.}\)

\(\begin{rcases}-9 x + 7 y = c \\ \text{door } A (-8 , -1)\end{rcases} c = -9 ⋅ -8 + 7 ⋅ -1 = 65\)
Dus \(l{:}\,-9 x + 7 y = 65 \text{.}\)

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

Aftrekken geeft \(2 x = 5\) dus \(x = 2\frac{1}{2} \text{.}\)

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

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

Substitutie geeft \(3 x - 2 (-3 x - 1) = 5\)

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

\(k{:}\,x - 7 y = 9\) omschrijven geeft \(y = \frac{1}{7} x - 1\frac{2}{7}\) dus \(\text{rc}_{k} = \frac{1}{7} \text{.}\)
\(l{:}\,8 x - 4 y = 6\) omschrijven geeft \(y = 2 x - 1\frac{1}{2}\) dus \(\text{rc}_{l} = 2 \text{.}\)

\(\tan(\alpha ) = \frac{1}{7}\) geeft \(\alpha = \tan^{-1}(\frac{1}{7}) = 8{,}13...\degree \text{.}\)
\(\tan(\beta ) = 2\) geeft \(\beta = \tan^{-1}(2) = 63{,}43...\degree \text{.}\)

\(\varphi = \alpha - \beta = 8{,}13...\degree - 63{,}43...\degree = -55{,}30...\degree \text{,}\) dus de gevraagde hoek is \(55{,}3\degree \text{.}\)

\(-\frac{6}{18} ≠ -\frac{3}{2} ≠ \frac{4}{5} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({3 \over 9} = {-6 \over p} = {q \over 12}\)

\({3 \over 9} = {-6 \over p}\) geeft \(p = -18\) (en \({3 \over 9} = {q \over 12}\) geeft \(q = 4 \text{)}\)

Een snijpunt, dus \(p ≠ -18\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,-5 x + 4 y = c \text{.}\)

\(\begin{rcases}-5 x + 4 y = c \\ \text{door } A (6 , -1)\end{rcases} c = -5 ⋅ 6 + 4 ⋅ -1 = -34\)
Dus \(l{:}\,-5 x + 4 y = -34 \text{.}\)