How to do this question 9 regarding matrices and transformations?

enter image source here

1 Answer
Feb 12, 2018

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

Explanation:

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