Find the equation of the line that is perpendicular to the line $3x-2y=5$ and passes through the point $(1,2)$.

Step 1 ：1. The first step is to find the slope of the given line. In general, if a line's equation is in the form $Ax+By=C$, its slope is $-\frac{A}{B}\text{.}Inthiscase,thegivenlin{e}^{\prime }sequationis\text{(}3x-2y=5$, so its slope is $-\frac{3}{-2}=1.5$.

Step 2 ：2. The slope of a line perpendicular to a given line is the negative reciprocal of the given line's slope. Therefore, the slope of the line we are trying to find is $-\frac{1}{1.5}=-\frac{2}{3}$.

Step 3 ：3. Using the point-slope form of a linear equation ($y-y1=m(x-x1)$), where $m$ is the slope and $(x1,y1)$ is a point on the line, we can plug in the slope we found and the given point $(1,2)$ to find the equation of the line. This gives us $y-2=-\frac{2}{3}(x-1)$.

Step 4 ：4. Simplifying this equation gives $y=-\frac{2}{3}x+\frac{4}{3}+2$, or $y=-\frac{2}{3}x+\frac{10}{3}$.