\(\begin{cases}2 x + 3 y = -1 \\ 4 x + 5 y = -4\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}10 x + 15 y = -5 \\ 12 x + 15 y = -12\end{cases}\)

Aftrekken geeft \(-2 x = 7\) dus \(x = -3\frac{1}{2} \text{.}\)

\(\begin{rcases}2 x + 3 y = -1 \\ x = -3\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ -3\frac{1}{2} + 3 y = -1 \\ 3 y = 6 \\ y = 2\end{matrix}\)

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

Substitutie geeft \(4 x - 3 (2 x + 2) = -1\)

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

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

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

\(k{:}\,8 x - 4 y = 6\) omschrijven geeft \(y = 2 x - 1\frac{1}{2}\) dus \(\text{rc}_{k} = 2 \text{.}\)
\(l{:}\,7 x - 2 y = 5\) omschrijven geeft \(y = 3\frac{1}{2} x - 2\frac{1}{2}\) dus \(\text{rc}_{l} = 3\frac{1}{2} \text{.}\)

\(\tan(\alpha ) = 2\) geeft \(\alpha = \tan^{-1}(2) = 63{,}43...\degree \text{.}\)
\(\tan(\beta ) = 3\frac{1}{2}\) geeft \(\beta = \tan^{-1}(3\frac{1}{2}) = 74{,}05...\degree \text{.}\)

\(\varphi = \alpha - \beta = 63{,}43...\degree - 74{,}05...\degree = -10{,}61...\degree \text{,}\) dus de gevraagde hoek is \(10{,}6\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,-3 x + 4 y = c \text{.}\)

\(\begin{rcases}-3 x + 4 y = c \\ \text{door } A (-2 , 6)\end{rcases} c = -3 ⋅ -2 + 4 ⋅ 6 = 30\)
Dus \(l{:}\,-3 x + 4 y = 30 \text{.}\)

\(-\frac{6}{18} = -\frac{2}{6} ≠ \frac{1}{4} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

\({3 \over p} = {-5 \over -\frac{5}{4}} = {q \over -\frac{3}{2}}\)

\({3 \over p} = {-5 \over -\frac{5}{4}}\) geeft \(p = \frac{3}{4}\) (en \({-5 \over -\frac{5}{4}} = {q \over -\frac{3}{2}}\) geeft \(q = -6 \text{)}\)

Evenwijdig, dus \(p = \frac{3}{4}\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,8 x + 3 y = c \text{.}\)

\(\begin{rcases}8 x + 3 y = c \\ \text{door } A (5 , 6)\end{rcases} c = 8 ⋅ 5 + 3 ⋅ 6 = 58\)
Dus \(l{:}\,8 x + 3 y = 58 \text{.}\)