Finite Mathematics, Chapter 1, 1.1, Section 1.1, Problem 36

Determine $k$ so that the line through $(4,-1)$ and $(k,2)$ is
a.) parallel to $2x + 3y = 6$
b.) perpendicular t $5x - 2y = -1$

a.) If we transform the given line into point slope form, we have

$
\begin{equation}
\begin{aligned}
2x + 3y &= 6 \\
\\
3y &= -2x + 6 \\
\\
y &= -\frac{2}{3}x + 2
\end{aligned}
\end{equation}
$

Now that the line is in the slope intercept form $y = mx + b$. By observation, $\displaystyle m = -\frac{2}{3}$
And if the line is parallel to the line we are looking, then we can say that both lines have the same slope. Therefore,

$
\begin{equation}
\begin{aligned}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
\\
\frac{-2}{3} &= \frac{2-(-1)}{k - 4}\\
\\
\frac{-2}{3} &= \frac{2 + 1}{k - 4}\\
\\
-2(k - 4) &= 3(3)\\
\\
-2k + 8 &= 9 \\
\\
-2k &= 1\\
\\
-2k &= 1\\
\\
k &= -\frac{1}{2}
\end{aligned}
\end{equation}
$


b.) If we transform the given line into point slope form, we have

$
\begin{equation}
\begin{aligned}
5x - 2y &= -1\\
\\
-2y &= -5x - 1\\
\\
y &= \frac{-5x}{-2} - \frac{1}{-2}\\
\\
y &= \frac{5}{2} x + \frac{1}{2}
\end{aligned}
\end{equation}
$

Now that the line is in the slope intercept form $y = mx + b$. By observation, $\displaystyle m = \frac{5}{2}$. Therefore,
the slope of the perpendicular line is $\displaystyle m_{\perp} = -\frac{1}{\frac{5}{2}} = -\frac{2}{5}$. Hence, we have

$
\begin{equation}
\begin{aligned}
-\frac{2}{5} &= \frac{2 - (-1)}{k - 4}\\
\\
-\frac{2}{5} &= \frac{2 + 1}{k - 4}\\
\\
-2 (k - 4) &= 5(3)\\
\\
-2k + 8 &= 15\\
\\
-2k &= 7 \\
\\
k &= - \frac{7}{2}
\end{aligned}
\end{equation}
$