All points on the line y=-2x+8 will be transformed by the matrix bb(A)
Any points on this line will have coordinates of the form (k,-2k+8)
:.
((x'),(y'))=((0,-2),(-4,0))[((k),(-2k+8))+((-2),(2))]
color(white)(888888)=((0,-2),(-4,0))((k-2),(-2k+10))
color(white)(888888)=((4k-20),(-4k+8))
i.e.
x'=4k-20 and y'=-4k+8
Eliminating k:
k=(x+20)/4
y=-4((x+20)/4)+8
color(blue)(y=-x-12)
So all images of (x',y') lie on the line y=-x-12.
The line y=-x-12 is the image of y=-2x+8 under the transformation.
Note:
We could have found this line and alternate way. If we had generated two pairs of coordinates using bb(y=-2x+8), and used them in the transformation i.e for bb(X), we would have 2 pairs of coordinates of the image. This would enable us to find the equation of the line.
Example:
From y=-2x+8
For x=3 and x=4
y=2 and y=0color(white)(88) respectively.
Under the transformation we have:
x=-8 and y=-4
x=-4 and y=-8
(y_2-y_1)/(x_2-x_1)=(-8-(-4))/(-4-(-8))=-1
y-(-4)=-(x-(-8))
y+4=-x-8
y=-x-12color(white)(888) as expected.