All points on the line y=2x+3y=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