Parallelogram upper A upper B upper C upper D is graphed on a coordinate plane. The x-axis ranges from negative 9 to 10 in increments of 1. The y-axis ranges from negative 9 to 10 in increments of 1.
Suppose the parallelogram on the coordinate plane was dilated with respect to the origin using a magnitude of 12
. What would be the coordinates of vertex D′
?
To find the coordinates of vertex D', we need to dilate the coordinates of vertex D with respect to the origin using a magnitude of 12.
The coordinates of vertex D are (x, y).
To dilate a point (x, y) with respect to the origin using a magnitude of m, the coordinates of the dilated point (x', y') can be found using the formula:
x' = mx
y' = my
In this case, the magnitude is 12 and the coordinates of vertex D are (x, y). Therefore, the coordinates of vertex D' would be:
x' = 12x
y' = 12y
So, the coordinates of vertex D' would be (12x, 12y).
sorry thats wrong
I apologize for the incorrect response. To correctly find the coordinates of vertex D', we need to multiply each coordinate of vertex D by the magnitude of dilation, 12.
If the coordinates of vertex D are (x, y), then the coordinates of vertex D' would be (12x, 12y).
Please let me know if there is anything else I can help you with.
To find the coordinates of vertex D' after the dilation, we need to multiply the coordinates of vertex D by the magnitude of the dilation, 12.
The coordinates of vertex D are given as (x, y). To dilate with respect to the origin, we multiply both the x-coordinate and the y-coordinate by the magnitude of the dilation.
So, the coordinates of vertex D' would be (12*x, 12*y).