**Answer:**

#y=-3/5x-2#

**Explanation:**

For parallel lines, the slope is the same. Since this given line is already given in the form #y=mx+b#, we know that the slope of any parallel line will also be #m#. Here, #m=-3/5# from #y=-3/5 x+2#.

The question is asking us to find an equation of the line that passes through the point #(0,-2)#. Well, this means that our line will have a point where #x=0# and #y=-2#, but with a given slope of #m=-3/5#. We just need to find a #b# such that the point #(0,-2)# exists on that line.

We can do this in two ways.

1) Use #y=mx+b# with #x=0#, #y=-2#, and #m=-3/5# to solve #b# and find this general equation of the parallel line.

Let's plug in our values.

#-2=(-3/5)(0)+b#

#b=-2#

We now know our #b#, which we can plug into the new #y=mx+b# (of which we already know #m#, since it remains the same). So, the equation of the line parallel to #y=-3/5 x+2# that goes through the point #(0,-2)# is

#y=-3/5x-2#

OR

2) Use the point-slope form of #y-y_1=m(x-x_1)#, where #y_1# and #x_1#=coordinates of point on line and #m#=slope, to find the equation directly. Here, #x_1=0# and #y_1=-2#, from #(0,-2)# and #m# is still #-3/5#.

Let's solve to find the equation of the line.

#y-y_1=m(x-x_1)#

#y-(-2)=-3/5(x-0)#

#y+2=-3/5x#

#y=-3/5x-2#