Voor het snijpunt met de \(x \text{-}\)as geldt \(y = 0 \text{,}\)
\(9 x + 8 ⋅ 0 = 24\) geeft \(x = 2\frac{2}{3} \text{,}\) dus \((2\frac{2}{3} , 0) \text{.}\)

Voor het snijpunt met de \(y \text{-}\)as geldt \(x = 0 \text{,}\)
\(9 ⋅ 0 + 8 y = 24\) geeft \(y = 3 \text{,}\) dus \((0 , 3) \text{.}\)

\(A (9 , -8\frac{4}{5})\) invullen geeft \(6 ⋅ 9 + 5 ⋅ -8\frac{4}{5} = 10 ≠ 7\)
Klopt niet, dus \(A\) ligt niet op \(l \text{.}\)

Herleiden geeft
\(7 x - 4 y = 8\)
\(-4 y = -7 x + 8\)
\(y = 1\frac{3}{4} x - 2 \text{.}\)

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

Vermenigvuldigen met \(4\) geeft
\(12 x + 4 y = 3 \text{.}\)

\(\begin{rcases}-8 x - 4 y = 76 \\ \text{door } A (-5 , a)\end{rcases} \begin{matrix}-8 ⋅ -5 - 4 ⋅ a = 76\end{matrix}\)

\(40 - 4 a = 76\)
\(-4 a = 36\)
\(a = -9 \text{.}\)

\(\begin{rcases}3 x - 7 y = 30 \\ (x , y) = (a , -6)\end{rcases} \begin{matrix}3 ⋅ a - 7 ⋅ -6 = 30\end{matrix}\)

\(3 a + 42 = 30\)
\(3 a = -12\)
\(a = -4 \text{.}\)

\(\begin{rcases}a x - 9 y = -37 \\ \text{door } A (2 , 5)\end{rcases} \begin{matrix}a ⋅ 2 - 9 ⋅ 5 = -37\end{matrix}\)

\(2 a - 45 = -37\)
\(2 a = 8\)
\(a = 4 \text{.}\)

\(\begin{rcases}2 x - 9 y = c \\ \text{door } A (5 , -7)\end{rcases} \begin{matrix}2 ⋅ 5 - 9 ⋅ -7 = c\end{matrix}\)

\(c = 10 + 63 = 73 \text{.}\)

Herleiden naar \(y = a x + b\) geeft
\(9 x + 7 y = -5\)
\(7 y = -9 x - 5\)
\(y = -1\frac{2}{7} x - \frac{5}{7} \text{.}\)

Dus \(\text{rc}_{l} = -1\frac{2}{7} \text{.}\)