For this case we have the following equation:
[tex]4x-8y = 9[/tex]
We manipulate algebraically to bring the equation to the slope-intersection form y = mx + b
Where:
m: It's the slope
b: It is the cut-off point with the y axis
[tex]8y = 4x-9\\y =\frac {4} {8} x- \frac {9} {8}\\y =\frac {1} {2} x- \frac {9} {8}[/tex]
By definition, if two lines are parallel then their slopes are equal.
Thus, the line is of the form:
[tex]y = \frac {1} {2} x + b[/tex]
We find the cut-off point by replacing the given point:
[tex]6 = \frac {1} {2} (- 3) + b\\6 = - \frac {3} {2} + b\\b = 6 + \frac {3} {2}\\b = \frac {12 + 3} {2}\\b = \frac {15} {2}[/tex]
Thus, the line is of the form:
[tex]y = \frac {1} {2} x + \frac {15} {2}[/tex]
Answer:
[tex]y = \frac {1} {2} x + \frac {15} {2}[/tex]
