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.

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:

Midpoint of AC:

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

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)

the answer is. -82562234434584

step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°