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

\(\tan(\alpha ) = 8\) geeft \(\alpha = \tan^{-1}(8) = 82{,}87...\degree \text{.}\)
\(\tan(\beta ) = -\frac{4}{7}\) geeft \(\beta = \tan^{-1}(-\frac{4}{7}) = -29{,}74...\degree \text{.}\)

\(\varphi = \alpha - \beta = 82{,}87...\degree - -29{,}74...\degree = 112{,}61...\degree \text{,}\) dus de gevraagde hoek is \(180\degree - 112{,}61...\degree = 67{,}4\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,-6 x - 2 y = c \text{.}\)

\(\begin{rcases}-6 x - 2 y = c \\ \text{door } A (-9 , -1)\end{rcases} c = -6 ⋅ -9 - 2 ⋅ -1 = 56\)
Dus \(l{:}\,-6 x - 2 y = 56 \text{.}\)

\(\frac{1}{2} = \frac{3}{6} ≠ -\frac{4}{5} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

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

Aftrekken geeft \(6 x = -6\) dus \(x = -1 \text{.}\)

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

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

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

\({5 \over 15} = {p \over 12} = {-1 \over q}\)

\({5 \over 15} = {p \over 12}\) geeft \(p = 4\) (en \({5 \over 15} = {-1 \over q}\) geeft \(q = -3 \text{)}\)

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

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

\(\begin{rcases}-4 x + y = c \\ \text{door } A (7 , -3)\end{rcases} c = -4 ⋅ 7 + 1 ⋅ -3 = -31\)
Dus \(l{:}\,-4 x + y = -31 \text{.}\)