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

Aftrekken geeft \(-3 x = 9\) dus \(x = -3 \text{.}\)

\(\begin{rcases}x - 2 y = 4 \\ x = -3\end{rcases} \begin{matrix}1 ⋅ -3 - 2 y = 4 \\ -2 y = 7 \\ y = -3\frac{1}{2}\end{matrix}\)

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

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

\(2 x - 4 x - 6 = 1\)
\(-2 x = 7\)
Dus \(x = -3\frac{1}{2} \text{.}\)

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

Dus \(S (-3\frac{1}{2} , 4) \text{.}\)

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

\(\tan(\alpha ) = 6\) geeft \(\alpha = \tan^{-1}(6) = 80{,}53...\degree \text{.}\)
\(\tan(\beta ) = -2\frac{1}{2}\) geeft \(\beta = \tan^{-1}(-2\frac{1}{2}) = -68{,}19...\degree \text{.}\)

\(\varphi = \alpha - \beta = 80{,}53...\degree - -68{,}19...\degree = 148{,}73...\degree \text{,}\) dus de gevraagde hoek is \(180\degree - 148{,}73...\degree = 31{,}3\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,-4 x + 9 y = c \text{.}\)

\(\begin{rcases}-4 x + 9 y = c \\ \text{door } A (1 , -5)\end{rcases} c = -4 ⋅ 1 + 9 ⋅ -5 = -49\)
Dus \(l{:}\,-4 x + 9 y = -49 \text{.}\)

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

\({1 \over 2} = {p \over -6} = {q \over 12}\)

\({1 \over 2} = {p \over -6}\) geeft \(p = -3\) en \({1 \over 2} = {q \over 12}\) geeft \(q = 6\)

Samenvallen, dus \(p = -3\) en \(q = 6 \text{.}\)

\(k \perp l \text{,}\) dus \(l{:}\,-9 x - 6 y = c \text{.}\)

\(\begin{rcases}-9 x - 6 y = c \\ \text{door } A (4 , 5)\end{rcases} c = -9 ⋅ 4 - 6 ⋅ 5 = -66\)
Dus \(l{:}\,-9 x - 6 y = -66 \text{.}\)