With a scale factor of 0.5 and the center of dilation at (0,0), what are the coordinates of the points of the image P' Q' R' S'?
(Note that P=2,2; Q=6,4; R=6,8; S=2,8)
in general if you have a point P that you are going to scale about a center of dilation (cx,cy) you end up with
$$P'=sf((P_x-c_x), (P_y-c_y))+(c_x,c_y)=$$
$$(sf(P_x-c_x)+c_x,~sf(P_y-c_y)+c_y)$$
here
$$c_x=c_y=0,~~sf=0.5$$
so we end up with
$$P'=(0.5 P_x, 0.5 P_y)$$
applying that to P=(2,2) we get
$$P'=(0.5\cdot 2, 0.5 \cdot 2)=(1,1)$$
you can figure out the rest of them
in general if you have a point P that you are going to scale about a center of dilation (cx,cy) you end up with
$$P'=sf((P_x-c_x), (P_y-c_y))+(c_x,c_y)=$$
$$(sf(P_x-c_x)+c_x,~sf(P_y-c_y)+c_y)$$
here
$$c_x=c_y=0,~~sf=0.5$$
so we end up with
$$P'=(0.5 P_x, 0.5 P_y)$$
applying that to P=(2,2) we get
$$P'=(0.5\cdot 2, 0.5 \cdot 2)=(1,1)$$
you can figure out the rest of them
(Note that P=2,2; Q=6,4; R=6,8; S=2,8)
Hi Rom,
The wording of this question is not familiar to me, nor do i understand your answer, but, if a shape is being shrunk by 50% about (0,0) isn't it just a matter of halving all the x values and halving all the y values hence getting
P'(1,1), Q'(3,2) R'(3,4) S'(1,4) ?
You are right Melody; but what if the centre of dilation was, say (2,3)? In this case you would need to halve the difference between the Px-values and 2, and the difference between the Py-values and 3, and then add these halved differences back on to 2 and 3 respectively to get the P'x and P'y coordinates. Rom gave the general procedure first, then started to apply it to the particular data set of the question.