\(\begin{cases}4 x + 2 y = 3 \\ 3 x + 4 y = 1\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}8 x + 4 y = 6 \\ 3 x + 4 y = 1\end{cases}\)

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

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

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

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

\(3 x - 6 x - 2 = -4\)
\(-3 x = -2\)
Dus \(x = \frac{2}{3} \text{.}\)

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

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

\(k{:}\,9 x - 2 y = 7\) omschrijven geeft \(y = 4\frac{1}{2} x - 3\frac{1}{2}\) dus \(\text{rc}_{k} = 4\frac{1}{2} \text{.}\)
\(l{:}\,x - 6 y = 5\) omschrijven geeft \(y = \frac{1}{6} x - \frac{5}{6}\) dus \(\text{rc}_{l} = \frac{1}{6} \text{.}\)

\(\tan(\alpha ) = 4\frac{1}{2}\) geeft \(\alpha = \tan^{-1}(4\frac{1}{2}) = 77{,}47...\degree \text{.}\)
\(\tan(\beta ) = \frac{1}{6}\) geeft \(\beta = \tan^{-1}(\frac{1}{6}) = 9{,}46...\degree \text{.}\)

\(\varphi = \alpha - \beta = 77{,}47...\degree - 9{,}46...\degree = 68{,}00...\degree \text{,}\) dus de gevraagde hoek is \(68{,}0\degree \text{.}\)

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

\(\begin{rcases}5 x - y = c \\ \text{door } A (6 , -8)\end{rcases} c = 5 ⋅ 6 - 1 ⋅ -8 = 38\)
Dus \(l{:}\,5 x - y = 38 \text{.}\)

\(\frac{3}{9} ≠ -\frac{6}{2} ≠ \frac{1}{5} \text{,}\) dus de lijnen \(k\) en \(l\) snijden.

\({p \over \frac{5}{2}} = {2 \over 1} = {q \over \frac{3}{2}}\)

\({p \over \frac{5}{2}} = {2 \over 1}\) geeft \(p = 5\) (en \({2 \over 1} = {q \over \frac{3}{2}}\) geeft \(q = 3 \text{)}\)

Een snijpunt, dus \(p ≠ 5\) en \(q\) mag elk getal zijn.

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

\(\begin{rcases}-x + 3 y = c \\ \text{door } A (8 , -7)\end{rcases} c = -1 ⋅ 8 + 3 ⋅ -7 = -29\)
Dus \(l{:}\,-x + 3 y = -29 \text{.}\)