I need the answers for this question. I am confused

Step-by-step explanation:

As C and D are both at y = -9, the line below V₂ is vertical and located at

x = ½(22 + (-2) = 10

The line between V₁ and V₂ will contain the point halfway between B and D.

x = ½(14 + (-2)) = 6

y = ½(7 + (-9)) = -1

(6, -1)

The slope of the line between V₁ and V₂ is the same as between V₁ and our just found midpoint between B and D

m = (5 - (-1)) / (0 - 6) = -1

so the line equation is

y = -x + 5

Which intersects x = 10 at

y = -10 + 5 = -5

V₂ = (10, -5)

The distance between V₂ and B, C, and D is

d = √((14 - 10)² + (7 - (-5))²) = √160

The distance between V₁ and B, A, and D is

d = √((-10 - 0)² + (15 - 5)²) = √200

As √200 > √160, the dump should be located at V₁ or at least √200 away from any two of the points

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'.