Question

Square ABCD is located on a coordinate plane. The coordinates for three of the vertices are listed below. ~A(2,7) ~C(8,1) ~D(2,1) Square ABCD is dilated by a scale factor of 2 with the center of dilation at the origin, to form square A' B' C' D'. What are the coordinates of vertex B'? Explain how you determined your answer.

Answer 1

Answer:

B' (16, 14)

Step-by-step explanation:

Whenever we have a dilation with a scale factor b, we get the new point as:

Let original point be (x,y), then the dilation would make it (bx, by).

There is a diagonal that goes through AC and another through BD. The midpoint of both the diagonal is same. Midpoint formula is:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Midpoint of AC:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\=(\frac{2+8}{2},\frac{7+1}{2})\\=(5,4)$

Now let vertex B have coordinates (x,y). Midpoint of BD:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\=(\frac{x+2}{2},\frac{y+1}{2})$

Equate this to the midpoint of AC:

(x+2)/2=5

x+2 = 10

x = 8

and

(y+1)/2=4

y+1=8

y = 7

Thus, B(8,7)

recalling dilation of scale factor 2 rule we said at the beginning, B' would be twice of each coordinate. Hence B' = (16,14)