Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and rounded to the nearest tenth. A (−9, 3), B (−6, −1), C (2, 5), D (−5, 6)

Perimeter = 5 + 5 + 7.1 + 10 =27.1

area = (5 + 10)x5  / 2 = 37.5

Step-by-step explanation:

Slope formula: m = (x' - x)/(y' - y)

m-AD: (6-3)/(-5 - -9) = 3/4    (I'm not detail other Other slopes)

m- BC = 3/4

m-AB = - 4/3

m-AD = m-BC         AD // BC      (ABCD is a Trapezoid)

m-AB = - 1 / m-BC

AB ⊥ BC

segment length formula: d = √(x' - x)² + (y' - y)²

AD = √(-5 - -9)² + (6 - 3)² = √4² + 3² = √25 = 5

DC = √50 = 7.1

BC = √100 = 10

AB = √25 = 5

Perimeter = 5 + 5 + 7.1 + 10 =27.1

area = (5 + 10)x5  / 2 = 37.5

area ABCD = (5+10)x5  / 2 =

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval