\(k \parallel l \text{,}\) dus \(l{:}\,9 x - 2 y = c \text{.}\)

\(\begin{rcases}9 x - 2 y = c \\ \text{door } A (6 , 8)\end{rcases} c = 9 ⋅ 6 - 2 ⋅ 8 = 38\)
Dus \(l{:}\,9 x - 2 y = 38 \text{.}\)

\(\begin{cases}4 x - 3 y = -1 \\ 2 x - 4 y = -3\end{cases}\) \(\begin{vmatrix}4 \\ 3\end{vmatrix}\) geeft \(\begin{cases}16 x - 12 y = -4 \\ 6 x - 12 y = -9\end{cases}\)

Aftrekken geeft \(10 x = 5\) dus \(x = \frac{1}{2} \text{.}\)

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

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

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

\(4 x - 6 x - 9 = -2\)
\(-2 x = 7\)
Dus \(x = -3\frac{1}{2} \text{.}\)

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

Dus \(S (-3\frac{1}{2} , 4) \text{.}\)

\(k{:}\,9 x - 2 y = -1\) omschrijven geeft \(y = 4\frac{1}{2} x + \frac{1}{2}\) dus \(\text{rc}_{k} = 4\frac{1}{2} \text{.}\)
\(l{:}\,7 x + 3 y = 8\) omschrijven geeft \(y = -2\frac{1}{3} x + 2\frac{2}{3}\) dus \(\text{rc}_{l} = -2\frac{1}{3} \text{.}\)

\(\tan(\alpha ) = 4\frac{1}{2}\) geeft \(\alpha = \tan^{-1}(4\frac{1}{2}) = 77{,}47...\degree \text{.}\)
\(\tan(\beta ) = -2\frac{1}{3}\) geeft \(\beta = \tan^{-1}(-2\frac{1}{3}) = -66{,}80...\degree \text{.}\)

\(\varphi = \alpha - \beta = 77{,}47...\degree - -66{,}80...\degree = 144{,}27...\degree \text{,}\) dus de gevraagde hoek is \(180\degree - 144{,}27...\degree = 35{,}7\degree \text{.}\)

\(-\frac{1}{2} = -\frac{3}{6} = -\frac{4}{8} \text{,}\) dus de lijnen \(k\) en \(l\) vallen samen.

\({2 \over 6} = {4 \over p} = {q \over -15}\)

\({2 \over 6} = {4 \over p}\) geeft \(p = 12\) en \({2 \over 6} = {q \over -15}\) geeft \(q = -5\)

Geen punt gemeenschappelijk, dus \(p = 12\) en \(q ≠ -5 \text{.}\)

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

\(\begin{rcases}8 x + 9 y = c \\ \text{door } A (6 , 4)\end{rcases} c = 8 ⋅ 6 + 9 ⋅ 4 = 84\)
Dus \(l{:}\,8 x + 9 y = 84 \text{.}\)