PL HELP THIS DUE TODAY BEFORE 8:00 .(05.05)The width of a rectangle is shown below: A coordinate plane with a point A at negative 3, 3 and C at negative 3, negative 2.

If the area of the rectangle to be drawn is 20 square units, where should points B and D be located if they lie to the right of points A and C?

B(1, 3) and D(1, −2)
B(3, 3) and D(3, −2)
B(2, 3) and D(2, −2)
B(1, 2) and D(1, −3)

the first option (1,3) and (1,-2)

Step-by-step explanation:

it is unusual to have neighboring corners of a rectangle called A and C. normally the neighbors of A are B and D. but ok.

A and C have the same x coordinate (-3). so, A is directly above C. the distance between them is therefore only the difference between their y coordinates.

AC = 3 - -2 = 3 + 2 = 5

the area of the rectangle is length × width, so

20 = AC × AB = 5 × AB

AB = 20/5 = 4

in a rectangle all angles are 90 degrees.

so, to find B we need to draw a line of 4 units from A to the right, perpendicular to AC. that means that now the y coordinate is the same as for A (3), and the distance to A is just the difference in the x coordinate :

4 = Bx - -3 = Bx + 3

Bx = 1

=> B = (1, 3)

to maintain the rectangle shape now D has to be right under B and to the right of C with the same distances

DB = AC = 5

DC = AB = 4

due to the special orientation of the rectangle (the sides are all parallel to the x- and y-axis) we can simply add the distances to the corresponding coordinates, or just deduct from the other points :

D has to have the same x coordinate as B (1).

D has to have the same y coordinate as C (-2).

so, D = (1, -2)

the line of

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)