Answer:
y=-3/5x-2y=−35x−2
Explanation:
For parallel lines, the slope is the same. Since this given line is already given in the form y=mx+by=mx+b, we know that the slope of any parallel line will also be mm. Here, m=-3/5m=−35 from y=-3/5 x+2y=−35x+2.
The question is asking us to find an equation of the line that passes through the point (0,-2)(0,−2). Well, this means that our line will have a point where x=0x=0 and y=-2y=−2, but with a given slope of m=-3/5m=−35. We just need to find a bb such that the point (0,-2)(0,−2) exists on that line.
We can do this in two ways.
1) Use y=mx+by=mx+b with x=0x=0, y=-2y=−2, and m=-3/5m=−35 to solve bb and find this general equation of the parallel line.
Let's plug in our values.
-2=(-3/5)(0)+b−2=(−35)(0)+b
b=-2b=−2
We now know our bb, which we can plug into the new y=mx+by=mx+b (of which we already know mm, since it remains the same). So, the equation of the line parallel to y=-3/5 x+2y=−35x+2 that goes through the point (0,-2)(0,−2) is
y=-3/5x-2y=−35x−2
OR
2) Use the point-slope form of y-y_1=m(x-x_1)y−y1=m(x−x1), where y_1y1 and x_1x1=coordinates of point on line and mm=slope, to find the equation directly. Here, x_1=0x1=0 and y_1=-2y1=−2, from (0,-2)(0,−2) and mm is still -3/5−35.
Let's solve to find the equation of the line.
y-y_1=m(x-x_1)y−y1=m(x−x1)
y-(-2)=-3/5(x-0)y−(−2)=−35(x−0)
y+2=-3/5xy+2=−35x
y=-3/5x-2y=−35x−2
