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

\(\begin{rcases}-2 x + 3 y = c \\ \text{door } A (-8 , 9)\end{rcases} c = -2 ⋅ -8 + 3 ⋅ 9 = 43\)
Dus \(l{:}\,-2 x + 3 y = 43 \text{.}\)

\(\begin{cases}3 x + 2 y = -4 \\ 5 x - 4 y = -3\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}6 x + 4 y = -8 \\ 5 x - 4 y = -3\end{cases}\)

Optellen geeft \(11 x = -11\) dus \(x = -1 \text{.}\)

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

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

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

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

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

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

\(k{:}\,7 x + 3 y = -5\) omschrijven geeft \(y = -2\frac{1}{3} x - 1\frac{2}{3}\) dus \(\text{rc}_{k} = -2\frac{1}{3} \text{.}\)
\(l{:}\,x + 2 y = -8\) omschrijven geeft \(y = -\frac{1}{2} x - 4\) dus \(\text{rc}_{l} = -\frac{1}{2} \text{.}\)

\(\tan(\alpha ) = -2\frac{1}{3}\) geeft \(\alpha = \tan^{-1}(-2\frac{1}{3}) = -66{,}80...\degree \text{.}\)
\(\tan(\beta ) = -\frac{1}{2}\) geeft \(\beta = \tan^{-1}(-\frac{1}{2}) = -26{,}56...\degree \text{.}\)

\(\varphi = \alpha - \beta = -66{,}80...\degree - -26{,}56...\degree = -40{,}23...\degree \text{,}\) dus de gevraagde hoek is \(40{,}2\degree \text{.}\)

\(-\frac{4}{12} ≠ -\frac{5}{2} ≠ \frac{3}{6} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({6 \over 18} = {p \over 3} = {5 \over q}\)

\({6 \over 18} = {p \over 3}\) geeft \(p = 1\) en \({6 \over 18} = {5 \over q}\) geeft \(q = 15\)

Geen punt gemeenschappelijk, dus \(p = 1\) en \(q ≠ 15 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,7 x + 2 y = c \text{.}\)

\(\begin{rcases}7 x + 2 y = c \\ \text{door } A (-4 , -3)\end{rcases} c = 7 ⋅ -4 + 2 ⋅ -3 = -34\)
Dus \(l{:}\,7 x + 2 y = -34 \text{.}\)