\(k{:}\,7 x - 6 y = -5\) omschrijven geeft \(y = 1\frac{1}{6} x + \frac{5}{6}\) dus \(\text{rc}_{k} = 1\frac{1}{6} \text{.}\)
\(l{:}\,x - 8 y = 9\) omschrijven geeft \(y = \frac{1}{8} x - 1\frac{1}{8}\) dus \(\text{rc}_{l} = \frac{1}{8} \text{.}\)

\(\tan(\alpha ) = 1\frac{1}{6}\) geeft \(\alpha = \tan^{-1}(1\frac{1}{6}) = 49{,}39...\degree \text{.}\)
\(\tan(\beta ) = \frac{1}{8}\) geeft \(\beta = \tan^{-1}(\frac{1}{8}) = 7{,}12...\degree \text{.}\)

\(\varphi = \alpha - \beta = 49{,}39...\degree - 7{,}12...\degree = 42{,}27...\degree \text{,}\) dus de gevraagde hoek is \(42{,}3\degree \text{.}\)

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

\(\begin{rcases}-4 x - 6 y = c \\ \text{door } A (-7 , -2)\end{rcases} c = -4 ⋅ -7 - 6 ⋅ -2 = 40\)
Dus \(l{:}\,-4 x - 6 y = 40 \text{.}\)

\(\frac{5}{10} = \frac{4}{8} = \frac{6}{12} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

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

Aftrekken geeft \(-6 x = 9\) dus \(x = -1\frac{1}{2} \text{.}\)

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

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

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

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

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

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

\({6 \over 18} = {1 \over p} = {-4 \over q}\)

\({6 \over 18} = {1 \over p}\) geeft \(p = 3\) (en \({6 \over 18} = {-4 \over q}\) geeft \(q = -12 \text{)}\)

Een snijpunt, dus \(p ≠ 3\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,-7 x - 4 y = c \text{.}\)

\(\begin{rcases}-7 x - 4 y = c \\ \text{door } A (3 , -1)\end{rcases} c = -7 ⋅ 3 - 4 ⋅ -1 = -17\)
Dus \(l{:}\,-7 x - 4 y = -17 \text{.}\)