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

\(\begin{rcases}-x + 3 y = c \\ \text{door } A (-2 , -7)\end{rcases} c = -1 ⋅ -2 + 3 ⋅ -7 = -19\)
Dus \(l{:}\,-x + 3 y = -19 \text{.}\)

\(\begin{cases}4 x + 2 y = 1 \\ 3 x + 4 y = -3\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}8 x + 4 y = 2 \\ 3 x + 4 y = -3\end{cases}\)

Aftrekken geeft \(5 x = 5\) 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 \(3 x - 3 (3 x - 3) = -5\)

\(3 x - 9 x + 9 = -5\)
\(-6 x = -14\)
Dus \(x = 2\frac{1}{3} \text{.}\)

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

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

\(k{:}\,x + 9 y = -8\) omschrijven geeft \(y = -\frac{1}{9} x - \frac{8}{9}\) dus \(\text{rc}_{k} = -\frac{1}{9} \text{.}\)
\(l{:}\,6 x + 3 y = -5\) omschrijven geeft \(y = -2 x - 1\frac{2}{3}\) dus \(\text{rc}_{l} = -2 \text{.}\)

\(\tan(\alpha ) = -\frac{1}{9}\) geeft \(\alpha = \tan^{-1}(-\frac{1}{9}) = -6{,}34...\degree \text{.}\)
\(\tan(\beta ) = -2\) geeft \(\beta = \tan^{-1}(-2) = -63{,}43...\degree \text{.}\)

\(\varphi = \alpha - \beta = -6{,}34...\degree - -63{,}43...\degree = 57{,}09...\degree \text{,}\) dus de gevraagde hoek is \(57{,}1\degree \text{.}\)

\(\frac{5}{10} ≠ -\frac{3}{1} ≠ \frac{6}{4} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({2 \over p} = {3 \over 6} = {q \over 8}\)

\({2 \over p} = {3 \over 6}\) geeft \(p = 4\) (en \({3 \over 6} = {q \over 8}\) geeft \(q = 4 \text{)}\)

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

\(k \perp l \text{,}\) dus \(l{:}\,7 x - 9 y = c \text{.}\)

\(\begin{rcases}7 x - 9 y = c \\ \text{door } A (-2 , 6)\end{rcases} c = 7 ⋅ -2 - 9 ⋅ 6 = -68\)
Dus \(l{:}\,7 x - 9 y = -68 \text{.}\)