Voor het snijpunt met de \(x \text{-}\)as geldt \(y = 0 \text{,}\)
\(3 x + 8 ⋅ 0 = 8\) geeft \(x = 2\frac{2}{3} \text{,}\) dus \((2\frac{2}{3} , 0) \text{.}\)

Voor het snijpunt met de \(y \text{-}\)as geldt \(x = 0 \text{,}\)
\(3 ⋅ 0 + 8 y = 8\) geeft \(y = 1 \text{,}\) dus \((0 , 1) \text{.}\)

\(A (6 , -5\frac{7}{8})\) invullen geeft \(9 ⋅ 6 + 8 ⋅ -5\frac{7}{8} = 7 ≠ 4\)
Klopt niet, dus \(A\) ligt niet op \(l \text{.}\)

Herleiden geeft
\(-6 x - 4 y = 5\)
\(-6 x = 4 y + 5\)
\(x = -\frac{2}{3} y - \frac{5}{6} \text{.}\)

Uit \(y = 4 x - \frac{3}{4}\) volgt \(-4 x + y = -\frac{3}{4} \text{.}\)

Vermenigvuldigen met \(-4\) geeft
\(16 x - 4 y = 3 \text{.}\)

\(\begin{rcases}-9 x - 5 y = 19 \\ \text{door } A (a , 7)\end{rcases} \begin{matrix}-9 ⋅ a - 5 ⋅ 7 = 19\end{matrix}\)

\(-9 a - 35 = 19\)
\(-9 a = 54\)
\(a = -6 \text{.}\)

\(\begin{rcases}-5 x - 7 y = 41 \\ (x , y) = (a , -8)\end{rcases} \begin{matrix}-5 ⋅ a - 7 ⋅ -8 = 41\end{matrix}\)

\(-5 a + 56 = 41\)
\(-5 a = -15\)
\(a = 3 \text{.}\)

\(\begin{rcases}2 x + b y = -30 \\ \text{door } A (-9 , 4)\end{rcases} \begin{matrix}2 ⋅ -9 + b ⋅ 4 = -30\end{matrix}\)

\(-18 + 4 b = -30\)
\(4 b = -12\)
\(b = -3 \text{.}\)

\(\begin{rcases}-5 x - 7 y = c \\ \text{door } A (8 , -4)\end{rcases} \begin{matrix}-5 ⋅ 8 - 7 ⋅ -4 = c\end{matrix}\)

\(c = -40 + 28 = -12 \text{.}\)

Herleiden naar \(y = a x + b\) geeft
\(9 x + 8 y = 5\)
\(8 y = -9 x + 5\)
\(y = -1\frac{1}{8} x + \frac{5}{8} \text{.}\)

Dus \(\text{rc}_{l} = -1\frac{1}{8} \text{.}\)