Finite Mathematics, Chapter 1, 1.1, Section 1.1, Problem 32

Determine a equation in slope intercept form (where possible) for the line that goes through $(-2,6)$ and perpendicular to $2x - 3y = 5$

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

$
\begin{equation}
\begin{aligned}
2x - 3y &= 5 \\
\\
-3y &= -2x + 5 \\
\\
y &= \frac{-2}{-3} x + \frac{5}{-3}\\
\\
y &= \frac{2}{3} x - \frac{5}{3}
\end{aligned}
\end{equation}
$

Now that the line is in the slope intercept form $y = mx + b$. By observation, $\displaystyle m = \frac{2}{3}$
Thus, the slope of the perpendicular line is
$\displaystyle m_{\perp} = -\frac{3}{2}$. Thus,
By using the point slope form,the equation of the line will be $y - y_1 = m(x - x_1)$

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