\(\begin{cases}4 x + 2 y = -4 \\ 2 x + 5 y = 2\end{cases}\) \(\begin{vmatrix}5 \\ 2\end{vmatrix}\) geeft \(\begin{cases}20 x + 10 y = -20 \\ 4 x + 10 y = 4\end{cases}\)

Aftrekken geeft \(16 x = -24\) dus \(x = -1\frac{1}{2} \text{.}\)

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

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

Substitutie geeft \(2 x + 4 (-2 x - 2) = -5\)

\(2 x - 8 x - 8 = -5\)
\(-6 x = 3\)
Dus \(x = -\frac{1}{2} \text{.}\)

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

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

\(k{:}\,x - 4 y = -8\) omschrijven geeft \(y = \frac{1}{4} x + 2\) dus \(\text{rc}_{k} = \frac{1}{4} \text{.}\)
\(l{:}\,2 x + 7 y = 6\) omschrijven geeft \(y = -\frac{2}{7} x + \frac{6}{7}\) dus \(\text{rc}_{l} = -\frac{2}{7} \text{.}\)

\(\tan(\alpha ) = \frac{1}{4}\) geeft \(\alpha = \tan^{-1}(\frac{1}{4}) = 14{,}03...\degree \text{.}\)
\(\tan(\beta ) = -\frac{2}{7}\) geeft \(\beta = \tan^{-1}(-\frac{2}{7}) = -15{,}94...\degree \text{.}\)

\(\varphi = \alpha - \beta = 14{,}03...\degree - -15{,}94...\degree = 29{,}98...\degree \text{,}\) dus de gevraagde hoek is \(30{,}0\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,5 x - 8 y = c \text{.}\)

\(\begin{rcases}5 x - 8 y = c \\ \text{door } A (2 , 6)\end{rcases} c = 5 ⋅ 2 - 8 ⋅ 6 = -38\)
Dus \(l{:}\,5 x - 8 y = -38 \text{.}\)

\(-\frac{2}{6} = -\frac{6}{18} = -\frac{1}{3} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

\({p \over 8} = {2 \over 4} = {6 \over q}\)

\({p \over 8} = {2 \over 4}\) geeft \(p = 4\) (en \({2 \over 4} = {6 \over q}\) geeft \(q = 12 \text{)}\)

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

\(k \perp l \text{,}\) dus \(l{:}\,6 x + 9 y = c \text{.}\)

\(\begin{rcases}6 x + 9 y = c \\ \text{door } A (-7 , 2)\end{rcases} c = 6 ⋅ -7 + 9 ⋅ 2 = -24\)
Dus \(l{:}\,6 x + 9 y = -24 \text{.}\)