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
