Quadrilateral abcd is the result of a reflection of quadrilateral efgh over the line. which line segment in the image corresponds to eh in the pre-image? a.) ad b.) ab c.) bc d.) cd

A) AD

Step-by-step explanation:

Given: Quadrilateral ABCD is the result of a reflection of quadrilateral EFGH over the line.

We know that after reflection each point on the pre-image has a corresponding point on its image that is equidistant from a given line of reflection.

Since, EH shows the first and the last character of the name .

Thus, the line segment in the image corresponds to EH in the pre-image is AD because A is the first and D is the last character of the pre-image.

B

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h , k) are the coordinates of the centre and r is the radius

The midpoint of the diameter is the centre of the circle, hence

(h, k) = (2, 5)

The radius is the distance from the centre to a point on the circle

Calculate r using the distance formula

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (2, 5) and (x₂, y₂ ) = (- 3, - 7)

r =

 =

 = = = 13, hence

(x - 2)² + (y - 5)² = 13², thus

(x - 2)² + (y - 5)² = 169 → B