for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.
![m_ {1} * m_ {2} = - 1](/tex.php?f=m_ {1} * m_ {2} = - 1)
then, given the following line:
![6x-3y = 18\\6x-18 = 3y](/tex.php?f=6x-3y = 18\\6x-18 = 3y)
![\frac {6x} {3} - \frac {18} {3} = y\\y = \frac {6x} {3} - \frac {18} {3}\\y = 2x-6](/tex.php?f=\frac {6x} {3} - \frac {18} {3} = y\\y = \frac {6x} {3} - \frac {18} {3}\\y = 2x-6)
so, ![m_ {1} = 2](/tex.php?f=m_ {1} = 2)
we are looking for m_ {2}:
![m_{2}=\frac{-1}{m_{1}}\\m_{2}=\frac{-1}{2}\\m_{2}=-\frac{1}{2}](/tex.php?f=m_{2}=\frac{-1}{m_{1}}\\m_{2}=\frac{-1}{2}\\m_{2}=-\frac{1}{2})
then, the line is given by:
![y = - \frac {1} {2} x + b](/tex.php?f=y = - \frac {1} {2} x + b)
we find "b" replacing the given point:
![8 = - \frac {1} {2} (0) + b\\8 = b](/tex.php?f=8 = - \frac {1} {2} (0) + b\\8 = b)
finally, the equation is:
![y = - \frac {1} {2} x + 8](/tex.php?f=y = - \frac {1} {2} x + 8)
option c
![Me it is a geometry question and you](/tpl/images/05/00/VKBLjmouLV04XBbI.jpg)