For this case we have that by definition, the standard form of a linear equation is given by:
[tex]ax + by = c[/tex]
By definition, 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 point-slope equation of a line:
[tex]y+4 = -\frac {2} {3}(x-12)[/tex]
The slope is:
[tex]m_ {1} = - \frac {2} {3}[/tex]
We find the slope [tex]m_ {2}[/tex]of a perpendicular line:
[tex]m_ {2} = \frac {-1}{m_ {1}}\\m_ {2} = \frac {-1} {-\frac {2} {3}}\\m_ {2} = \frac{3} {2}[/tex]
Thus, the equation is of the form:
[tex]y-y_ {0} = \frac {3} {2} (x-x_ {0})[/tex]
We have the point through which the line passes:
[tex](x_ {0}, y_ {0}) :( 2, -3)[/tex]
Thus, the equation is:
[tex]y - (- 3) = \frac {3} {2} (x-2)\\y + 3 = \frac {3} {2} (x-2)[/tex]
We manipulate algebraically:
[tex]y + 3 = \frac{3} {2} x- \frac {3} {2} (2)\\y + 3 = \frac{3} {2} x-3[/tex]
We add 3 to both sides of the equation:
[tex]y + 3 + 3 = \frac {3} {2} x\\y + 6 = \frac {3} {2} x[/tex]
We multiply by 2 on both sides of the equation:
[tex]2(y + 6) = 3x\\2y + 12 = 3x[/tex]
We subtract 3x on both sides:
[tex]2y-3x + 12 = 0[/tex]
We subtract 12 from both sides:
[tex]2y-3x = -12[/tex]
ANswer:
[tex]-3x + 2y = -12[/tex]
