All points on the line # y=2x+3# will be transformed by the matrix #((0,3),(-2,0))#

Any point on the line will have coordinates of the form. #(k,2k+3)#

Thus:

#((x'),(y'))=((0,3),(-2,0))((k),(2k+3))#

#color(white)(888888)=((0+6k+9),(-2k+0))=((6k+9),(-2k))#

i.e. #x'=6k+9# **and** #y'=-2k#

Eliminating #k#

#k=(x-9)/6#

#y=-2((x-9)/6)#

#color(blue)(y=-1/3x+3)#

#y=-1/3x+3# is the image of #y=2x+3# under the transformation:

#((0,3),(-2,0))#

**Method 2**

Since we have:

#bb(AX)=bb(X^')#

Where #bb(A)=((0,3),(-2,0))#

#bb(X)=((x),(y))# and #bb(X^')=((x'),(y'))#

Generating a pair of coordinates using the line #y=2x+3#

#x=3# and #x=4#

gives:

#y=9# and #y=11# respectively.

Putting these under the transformation:

#((0,3),(-2,0))((x),(y))=((x'),(y'))#

#:.#

#((0,3),(-2,0))((3),(9))=((27),(-6))#

#((0,3),(-2,0))((4),(11))=((33),(-8))#

Gradient of image line:

#(y_2-y_1)/(x_2-x_1)=(-8-(-6))/(33-27)=(-2)/6=-1/3#

Using point slope form of a line:

#(y_2-y_1)=m(x_2-x_1)# #color(white)(88)m= "gradient"#

#y-(-8)=-1/3(x-33)#

#y+8=-1/3x+11#

#y=-1/3x+3color(white)(888)# **as expected**

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