\(\begin{cases}2 x - 3 y = 1 \\ 4 x - 5 y = 4\end{cases}\) \(\begin{vmatrix}5 \\ 3\end{vmatrix}\) geeft \(\begin{cases}10 x - 15 y = 5 \\ 12 x - 15 y = 12\end{cases}\)

Aftrekken geeft \(-2 x = -7\) dus \(x = 3\frac{1}{2} \text{.}\)

\(\begin{rcases}2 x - 3 y = 1 \\ x = 3\frac{1}{2}\end{rcases} \begin{matrix}2 ⋅ 3\frac{1}{2} - 3 y = 1 \\ -3 y = -6 \\ y = 2\end{matrix}\)

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

Substitutie geeft \(4 x + 1 (-2 x + 2) = 3\)

\(4 x - 2 x + 2 = 3\)
\(2 x = 1\)
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 \(S (\frac{1}{2} , 1) \text{.}\)

\(k{:}\,6 x + 8 y = -2\) omschrijven geeft \(y = -\frac{3}{4} x - \frac{1}{4}\) dus \(\text{rc}_{k} = -\frac{3}{4} \text{.}\)
\(l{:}\,4 x + 7 y = -5\) omschrijven geeft \(y = -\frac{4}{7} x - \frac{5}{7}\) dus \(\text{rc}_{l} = -\frac{4}{7} \text{.}\)

\(\tan(\alpha ) = -\frac{3}{4}\) geeft \(\alpha = \tan^{-1}(-\frac{3}{4}) = -36{,}86...\degree \text{.}\)
\(\tan(\beta ) = -\frac{4}{7}\) geeft \(\beta = \tan^{-1}(-\frac{4}{7}) = -29{,}74...\degree \text{.}\)

\(\varphi = \alpha - \beta = -36{,}86...\degree - -29{,}74...\degree = -7{,}12...\degree \text{,}\) dus de gevraagde hoek is \(7{,}1\degree \text{.}\)

\(k \parallel l \text{,}\) dus \(l{:}\,4 x + 7 y = c \text{.}\)

\(\begin{rcases}4 x + 7 y = c \\ \text{door } A (3 , 2)\end{rcases} c = 4 ⋅ 3 + 7 ⋅ 2 = 26\)
Dus \(l{:}\,4 x + 7 y = 26 \text{.}\)

\(\frac{2}{4} = \frac{1}{2} ≠ \frac{6}{3} \text{,}\) dus de lijnen \(k\) en \(l\) zijn evenwijdig.

\({p \over 12} = {5 \over 20} = {4 \over q}\)

\({p \over 12} = {5 \over 20}\) geeft \(p = 3\) en \({5 \over 20} = {4 \over q}\) geeft \(q = 16\)

Geen punt gemeenschappelijk, dus \(p = 3\) en \(q ≠ 16 \text{.}\)

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

\(\begin{rcases}-x - 2 y = c \\ \text{door } A (7 , 5)\end{rcases} c = -1 ⋅ 7 - 2 ⋅ 5 = -17\)
Dus \(l{:}\,-x - 2 y = -17 \text{.}\)