\(k \parallel l \text{,}\) dus \(l{:}\,8 x - y = c \text{.}\)

\(\begin{rcases}8 x - y = c \\ \text{door } A (-5 , -2)\end{rcases} c = 8 ⋅ -5 - 1 ⋅ -2 = -38\)
Dus \(l{:}\,8 x - y = -38 \text{.}\)

\(\begin{cases}4 x - 2 y = 0 \\ 2 x + y = 2\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 x - 2 y = 0 \\ 4 x + 2 y = 4\end{cases}\)

Optellen geeft \(8 x = 4\) dus \(x = \frac{1}{2} \text{.}\)

\(\begin{rcases}4 x - 2 y = 0 \\ x = \frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ \frac{1}{2} - 2 y = 0 \\ -2 y = -2 \\ y = 1\end{matrix}\)

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

Substitutie geeft \(2 x - 2 (2 x - 1) = -1\)

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

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

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

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

\(\tan(\alpha ) = 1\frac{1}{4}\) geeft \(\alpha = \tan^{-1}(1\frac{1}{4}) = 51{,}34...\degree \text{.}\)
\(\tan(\beta ) = -\frac{7}{9}\) geeft \(\beta = \tan^{-1}(-\frac{7}{9}) = -37{,}87...\degree \text{.}\)

\(\varphi = \alpha - \beta = 51{,}34...\degree - -37{,}87...\degree = 89{,}21...\degree \text{,}\) dus de gevraagde hoek is \(89{,}2\degree \text{.}\)

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

\({p \over 9} = {5 \over 15} = {-1 \over q}\)

\({p \over 9} = {5 \over 15}\) geeft \(p = 3\) (en \({5 \over 15} = {-1 \over q}\) geeft \(q = -3 \text{)}\)

Evenwijdig, dus \(p = 3\) en \(q\) mag elk getal zijn.

\(k \perp l \text{,}\) dus \(l{:}\,-x + 6 y = c \text{.}\)

\(\begin{rcases}-x + 6 y = c \\ \text{door } A (9 , -5)\end{rcases} c = -1 ⋅ 9 + 6 ⋅ -5 = -39\)
Dus \(l{:}\,-x + 6 y = -39 \text{.}\)