\(k \parallel l \text{,}\) dus \(l{:}\,-5 x + 2 y = c \text{.}\)

\(\begin{rcases}-5 x + 2 y = c \\ \text{door } A (1 , 3)\end{rcases} c = -5 ⋅ 1 + 2 ⋅ 3 = 1\)
Dus \(l{:}\,-5 x + 2 y = 1 \text{.}\)

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

Aftrekken geeft \(2 x = 5\) dus \(x = 2\frac{1}{2} \text{.}\)

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

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

Substitutie geeft \(3 x + 5 (-3 x + 3) = -1\)

\(3 x - 15 x + 15 = -1\)
\(-12 x = -16\)
Dus \(x = 1\frac{1}{3} \text{.}\)

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

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

\(k{:}\,2 x - 7 y = 5\) omschrijven geeft \(y = \frac{2}{7} x - \frac{5}{7}\) dus \(\text{rc}_{k} = \frac{2}{7} \text{.}\)
\(l{:}\,x + 6 y = -3\) omschrijven geeft \(y = -\frac{1}{6} x - \frac{1}{2}\) dus \(\text{rc}_{l} = -\frac{1}{6} \text{.}\)

\(\tan(\alpha ) = \frac{2}{7}\) geeft \(\alpha = \tan^{-1}(\frac{2}{7}) = 15{,}94...\degree \text{.}\)
\(\tan(\beta ) = -\frac{1}{6}\) geeft \(\beta = \tan^{-1}(-\frac{1}{6}) = -9{,}46...\degree \text{.}\)

\(\varphi = \alpha - \beta = 15{,}94...\degree - -9{,}46...\degree = 25{,}40...\degree \text{,}\) dus de gevraagde hoek is \(25{,}4\degree \text{.}\)

\(\frac{4}{8} = \frac{2}{4} ≠ -\frac{6}{3} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

\({1 \over 3} = {5 \over p} = {q \over -12}\)

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

Evenwijdig, dus \(p = 15\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,4 x + 2 y = c \text{.}\)

\(\begin{rcases}4 x + 2 y = c \\ \text{door } A (3 , -8)\end{rcases} c = 4 ⋅ 3 + 2 ⋅ -8 = -4\)
Dus \(l{:}\,4 x + 2 y = -4 \text{.}\)