\(\begin{cases}5 x - 4 y = -4 \\ 3 x - 2 y = -3\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}5 x - 4 y = -4 \\ 6 x - 4 y = -6\end{cases}\)

Aftrekken geeft \(-x = 2\) dus \(x = -2 \text{.}\)

\(\begin{rcases}5 x - 4 y = -4 \\ x = -2\end{rcases} \begin{matrix}5 ⋅ -2 - 4 y = -4 \\ -4 y = 6 \\ y = -1\frac{1}{2}\end{matrix}\)

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

Substitutie geeft \(2 x + 5 (-2 x + 3) = -5\)

\(2 x - 10 x + 15 = -5\)
\(-8 x = -20\)
Dus \(x = 2\frac{1}{2} \text{.}\)

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

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

\(k{:}\,2 x - 5 y = 1\) omschrijven geeft \(y = \frac{2}{5} x - \frac{1}{5}\) dus \(\text{rc}_{k} = \frac{2}{5} \text{.}\)
\(l{:}\,7 x + 3 y = -9\) omschrijven geeft \(y = -2\frac{1}{3} x - 3\) dus \(\text{rc}_{l} = -2\frac{1}{3} \text{.}\)

\(\tan(\alpha ) = \frac{2}{5}\) geeft \(\alpha = \tan^{-1}(\frac{2}{5}) = 21{,}80...\degree \text{.}\)
\(\tan(\beta ) = -2\frac{1}{3}\) geeft \(\beta = \tan^{-1}(-2\frac{1}{3}) = -66{,}80...\degree \text{.}\)

\(\varphi = \alpha - \beta = 21{,}80...\degree - -66{,}80...\degree = 88{,}60...\degree \text{,}\) dus de gevraagde hoek is \(88{,}6\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,6 x + 8 y = c \text{.}\)

\(\begin{rcases}6 x + 8 y = c \\ \text{door } A (3 , 4)\end{rcases} c = 6 ⋅ 3 + 8 ⋅ 4 = 50\)
Dus \(l{:}\,6 x + 8 y = 50 \text{.}\)

\(-\frac{4}{12} ≠ \frac{1}{6} ≠ -\frac{2}{3} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({4 \over 1} = {-5 \over p} = {-2 \over q}\)

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

Een snijpunt, dus \(p ≠ -1\frac{1}{4}\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,-8 x - 3 y = c \text{.}\)

\(\begin{rcases}-8 x - 3 y = c \\ \text{door } A (5 , -9)\end{rcases} c = -8 ⋅ 5 - 3 ⋅ -9 = -13\)
Dus \(l{:}\,-8 x - 3 y = -13 \text{.}\)