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.**

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