## College Algebra (10th Edition)

The equation of the line is $\color{blue}{y=-\frac{2}{3}x + \frac{17}{3}}$.
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 $y=-\frac{2}{3}x + \frac{17}{3}$