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 slope
b: It is the cut-off point with the y axis
In 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]
