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

Aftrekken geeft \(-3 x = 3\) dus \(x = -1 \text{.}\)

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

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

Substitutie geeft \(2 x - 1 (-2 x - 3) = 5\)

\(2 x + 2 x + 3 = 5\)
\(4 x = 2\)
Dus \(x = \frac{1}{2} \text{.}\)

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

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

\(k{:}\,3 x - 6 y = 4\) omschrijven geeft \(y = \frac{1}{2} x - \frac{2}{3}\) dus \(\text{rc}_{k} = \frac{1}{2} \text{.}\)
\(l{:}\,2 x + y = -5\) omschrijven geeft \(y = -2 x - 5\) dus \(\text{rc}_{l} = -2 \text{.}\)

\(\tan(\alpha ) = \frac{1}{2}\) geeft \(\alpha = \tan^{-1}(\frac{1}{2}) = 26{,}56...\degree \text{.}\)
\(\tan(\beta ) = -2\) geeft \(\beta = \tan^{-1}(-2) = -63{,}43...\degree \text{.}\)

\(\varphi = \alpha - \beta = 26{,}56...\degree - -63{,}43...\degree = 90\degree \text{,}\) dus de gevraagde hoek is \(90{,}0\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,9 x - 8 y = c \text{.}\)

\(\begin{rcases}9 x - 8 y = c \\ \text{door } A (4 , -7)\end{rcases} c = 9 ⋅ 4 - 8 ⋅ -7 = 92\)
Dus \(l{:}\,9 x - 8 y = 92 \text{.}\)

\(-\frac{5}{15} = -\frac{4}{12} ≠ \frac{2}{6} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

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

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

Evenwijdig, dus \(p = 1\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,3 x + 7 y = c \text{.}\)

\(\begin{rcases}3 x + 7 y = c \\ \text{door } A (4 , -5)\end{rcases} c = 3 ⋅ 4 + 7 ⋅ -5 = -23\)
Dus \(l{:}\,3 x + 7 y = -23 \text{.}\)