Answer
The equation of the line is $\color{blue}{y=-\frac{2}{3}x + \frac{17}{3}}$.
Work Step by Step
RECALL:
(1) The slope-intercept form of a line's equation is $y=mx+b$ where $m$ = slope and $b$ = y-intercept.
(2) Perpendicular lines have slopes whose product is $-1$ (slopes are negative reciprocals of each other).
Write the equation $3x-2y=7$ in slope-intercept form by isolating $y$ on the left side of the equation:
$3x-2y=7
\\-2y = -3x+7
\\\dfrac{-2y}{-2} = \dfrac{-3x+7}{-2}
\\y = \dfrac{3}{2}x - \dfrac{7}{2}$
The line we are looking for the equation of is perpendicular to the line above whose slope is $\dfrac{3}{2}$.
The negative reciprocal of $\dfrac{3}{2}$ is $-\dfrac{2}{3}$.
Thus, the slope of the line is $-\dfrac{2}{3}$ and its tentative equation is:
$y=-\dfrac{2}{3}x + b$.
Since the line contains the point, the coordinates of this point satisfies the equation of the line. Substitute the x and y coordinates of this point into the line's tentative equation to obtain:
$y=-\frac{2}{3}x+b
\\5=-\frac{2}{3}(1) + b
\\5=-\frac{2}{3} + b
\\5+\frac{2}{3} = b
\\\frac{15}{3} + \frac{2}{3} = b
\\\frac{17}{3} = b$
Therefore the equation of the line is $y=-\frac{2}{3}x + \frac{17}{3}$
