For this case we have by definition, that the equation of a line in the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cutoff point with the y axis
We have the following line:
[tex]2x-3y = 12\\2x-12 = 3y\\y = \frac {2} {3} x-4[/tex]
If the line we wish to find is perpendicular to the one given, then its slope is given by:
[tex]m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {\frac {2} {3}}\\m_ {2} = - \frac {3} {2}[/tex]
Then the line is:
[tex]y = - \frac {3} {2} x + b[/tex]
We substitute the point:
[tex]6 = - \frac {3} {2} (2) + b\\6 = -3 + b\\b = 6 + 3\\b = 9[/tex]
Finally, the equation is:
[tex]y = - \frac {3} {2} x + 9[/tex]
Answer:
[tex]y = - \frac {3} {2} x + 9[/tex]
