What are ways using coordinate geometry, that I could determine that this is a trapezoid?

One method to see it's a trapezoid is to find the slope of lines BC and AD.

The slope formula is

m = (y2-y1)/(x2-x1)

You should find that BC and AD both have the same slope (of -1), so that means the lines are parallel. That proves we have a trapezoid.

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To prove this trapezoid is isosceles, you can use the distance formula

to find the lengths of AB and CD (the two non-parallel sides). You should find that AB = CD.

Because AB and CD are horizontal and vertical respectively, this means you can simply count out the spaces to find that AB and CD are 3 units each. For any other rotated version of this trapezoid, use the distance formula instead.

found it xd it ok yt

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'.