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.
