\(k \parallel l \text{,}\) dus \(l{:}\,-x + 3 y = c \text{.}\)

\(\begin{rcases}-x + 3 y = c \\ \text{door } A (2 , -7)\end{rcases} c = -1 ⋅ 2 + 3 ⋅ -7 = -23\)
Dus \(l{:}\,-x + 3 y = -23 \text{.}\)

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

Aftrekken geeft \(2 x = 7\) dus \(x = 3\frac{1}{2} \text{.}\)

\(\begin{rcases}2 x - y = 2 \\ x = 3\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ 3\frac{1}{2} - y = 2 \\ -y = -5 \\ y = 5\end{matrix}\)

Dus \(S (3\frac{1}{2} , 5) \text{.}\)

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

\(3 x - 9 x + 6 = 4\)
\(-6 x = -2\)
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 \(S (\frac{1}{3} , -1) \text{.}\)

\(k{:}\,x + 5 y = -9\) omschrijven geeft \(y = -\frac{1}{5} x - 1\frac{4}{5}\) dus \(\text{rc}_{k} = -\frac{1}{5} \text{.}\)
\(l{:}\,2 x + 7 y = -8\) omschrijven geeft \(y = -\frac{2}{7} x - 1\frac{1}{7}\) dus \(\text{rc}_{l} = -\frac{2}{7} \text{.}\)

\(\tan(\alpha ) = -\frac{1}{5}\) geeft \(\alpha = \tan^{-1}(-\frac{1}{5}) = -11{,}30...\degree \text{.}\)
\(\tan(\beta ) = -\frac{2}{7}\) geeft \(\beta = \tan^{-1}(-\frac{2}{7}) = -15{,}94...\degree \text{.}\)

\(\varphi = \alpha - \beta = -11{,}30...\degree - -15{,}94...\degree = 4{,}63...\degree \text{,}\) dus de gevraagde hoek is \(4{,}6\degree \text{.}\)

\(\frac{1}{3} = \frac{3}{9} = \frac{4}{12} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

\({2 \over p} = {-5 \over -20} = {q \over 4}\)

\({2 \over p} = {-5 \over -20}\) geeft \(p = 8\) (en \({-5 \over -20} = {q \over 4}\) geeft \(q = 1 \text{)}\)

Evenwijdig, dus \(p = 8\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,5 x - 6 y = c \text{.}\)

\(\begin{rcases}5 x - 6 y = c \\ \text{door } A (1 , -9)\end{rcases} c = 5 ⋅ 1 - 6 ⋅ -9 = 59\)
Dus \(l{:}\,5 x - 6 y = 59 \text{.}\)