\(\begin{cases}2 x - 5 y = 4 \\ 4 x - 3 y = 1\end{cases}\) \(\begin{vmatrix}3 \\ 5\end{vmatrix}\) geeft \(\begin{cases}6 x - 15 y = 12 \\ 20 x - 15 y = 5\end{cases}\)

Aftrekken geeft \(-14 x = 7\) dus \(x = -\frac{1}{2} \text{.}\)

\(\begin{rcases}2 x - 5 y = 4 \\ x = -\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ -\frac{1}{2} - 5 y = 4 \\ -5 y = 5 \\ y = -1\end{matrix}\)

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

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

\(3 x - 15 x - 15 = 1\)
\(-12 x = 16\)
Dus \(x = -1\frac{1}{3} \text{.}\)

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

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

\(k{:}\,7 x - 3 y = -4\) omschrijven geeft \(y = 2\frac{1}{3} x + 1\frac{1}{3}\) dus \(\text{rc}_{k} = 2\frac{1}{3} \text{.}\)
\(l{:}\,x - 2 y = 6\) omschrijven geeft \(y = \frac{1}{2} x - 3\) 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 \(40{,}2\degree \text{.}\)

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

\(\begin{rcases}2 x + 7 y = c \\ \text{door } A (-5 , 1)\end{rcases} c = 2 ⋅ -5 + 7 ⋅ 1 = -3\)
Dus \(l{:}\,2 x + 7 y = -3 \text{.}\)

\(\frac{5}{15} = \frac{2}{6} = \frac{6}{18} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

\({4 \over 16} = {p \over -20} = {6 \over q}\)

\({4 \over 16} = {p \over -20}\) geeft \(p = -5\) en \({4 \over 16} = {6 \over q}\) geeft \(q = 24\)

Geen punt gemeenschappelijk, dus \(p = -5\) en \(q ≠ 24 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,-6 x - 8 y = c \text{.}\)

\(\begin{rcases}-6 x - 8 y = c \\ \text{door } A (-1 , -5)\end{rcases} c = -6 ⋅ -1 - 8 ⋅ -5 = 46\)
Dus \(l{:}\,-6 x - 8 y = 46 \text{.}\)