\(\begin{cases}3 x - 4 y = 3 \\ x - 2 y = 0\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}3 x - 4 y = 3 \\ 2 x - 4 y = 0\end{cases}\)

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

\(\begin{rcases}3 x - 4 y = 3 \\ x = 3\end{rcases} \begin{matrix}3 ⋅ 3 - 4 y = 3 \\ -4 y = -6 \\ y = 1\frac{1}{2}\end{matrix}\)

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

Substitutie geeft \(3 x - 5 (-3 x - 2) = 4\)

\(3 x + 15 x + 10 = 4\)
\(18 x = -6\)
Dus \(x = -\frac{1}{3} \text{.}\)

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

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

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

\(\tan(\alpha ) = 2\) geeft \(\alpha = \tan^{-1}(2) = 63{,}43...\degree \text{.}\)
\(\tan(\beta ) = -5\) geeft \(\beta = \tan^{-1}(-5) = -78{,}69...\degree \text{.}\)

\(\varphi = \alpha - \beta = 63{,}43...\degree - -78{,}69...\degree = 142{,}12...\degree \text{,}\) dus de gevraagde hoek is \(180\degree - 142{,}12...\degree = 37{,}9\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,6 x - 4 y = c \text{.}\)

\(\begin{rcases}6 x - 4 y = c \\ \text{door } A (-8 , -5)\end{rcases} c = 6 ⋅ -8 - 4 ⋅ -5 = -28\)
Dus \(l{:}\,6 x - 4 y = -28 \text{.}\)

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

\({2 \over 4} = {p \over -12} = {q \over 10}\)

\({2 \over 4} = {p \over -12}\) geeft \(p = -6\) en \({2 \over 4} = {q \over 10}\) geeft \(q = 5\)

Geen punt gemeenschappelijk, dus \(p = -6\) en \(q ≠ 5 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,-9 x - 5 y = c \text{.}\)

\(\begin{rcases}-9 x - 5 y = c \\ \text{door } A (-8 , -1)\end{rcases} c = -9 ⋅ -8 - 5 ⋅ -1 = 77\)
Dus \(l{:}\,-9 x - 5 y = 77 \text{.}\)