No line equation has been provided, but, using an example, we can see the method for finding a parallel line passing through the given point. Suppose the given line to be y=2x+1. The gradient of this line is 2 (coefficient of the x variable). The parallel line has the same gradient, so we can write y-9=2(x+3), which gives y=9 when x=-3.
This expands to y-9=2x+6, y=2x+15 and is the equation of the parallel line. This can also be written 2x-y+15=0.
Another example: given the line x+2y=7, 2y=7-x, y=7/2-x/2, which has the gradient -½.
So y-9=-(x+3)/2, 2y-18=-x-3, 2y=15-x and this can also be written x+2y=15.