Give the slope-intercept form of the equation of the line that is perpendicular to 5x + 2y = 12 and contains the point (2, 3).Hint: solve for y=mx+b in order to get the slope (m) and then substitute in the slope and the point to find b.

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:[tex]y = mx + b[/tex]Where:m: It's the slopeb: It is the cut-off point with the y axisIn addition, if two lines are perpendicular then the product of their slopes is -1. That is to say:[tex]m_ {1} * m_ {2} = - 1[/tex] We have the following line:[tex]5x + 2y = 12\\2y = -5x + 12\\y = - \frac {5} {2} x + 6[/tex]So, we have to:[tex]m_ {1} = - \frac {5} {2}[/tex]We find [tex]m_ {2}:[/tex][tex]m_ {2} = \frac {-1} {- \frac {5} {2}}\\m_ {2} = \frac {2} {5}[/tex]Therefore, the line is of the form:[tex]y = \frac {2} {5} x + b[/tex]We substitute the given point to find "b":[tex]3 = \frac {2} {5} (2) + b\\3 = \frac {4} {5} + b\\b = 3- \frac {4} {5}\\b = \frac {11} {5}[/tex]Finally, the equation is:[tex]y = \frac {2} {5} x + \frac {11} {5}[/tex]ANswer:[tex]y = \frac {2} {5} x + \frac {11} {5}[/tex]