Quadrilateral ABCD has a vertices A(-1,5) B(5,1) C(6,-2) D(x,2) what is the x coordinate to make ABCD a parallelogram

we know that
A Parallelogram is a flat shape with opposite sides parallel and equal in length

we have
A(-1,5) B(5,1) C(6,-2) D(x,2)

step 1
find the slope and the distance between point A and B
m=(y2-y1)/(x2-x1)----->m=(1-5)/(5+1) ------> m=-4/6----> m=-2/3
dAB=√[4²+6²----> dAB=√52 units

step 2
find the equation of the line CD parallel to the line AB and pass through the point C 
m=-2/3  (parallel lines have the same slope)
point C (6,-2)
y-y1=m*(x-x1)-------> y+2=(-2/3)*(x-6)----> y=(-2/3)x+4-2----> y=(-2/3)x+2

Step 3
with the equation of a line CD find the point D (x,2)
y=(-2/3)x+2
for y=2
2=(-2/3)x+2---> (-2/3)x=0-------> x=0
the point D is (0,2)

step 4
find the distance CD
C(6,-2) D(0,2)
d=√[(2+2)²+(6)²]--------> d=√52
so
dAB=dCD

step 5
find the slope and the distance between point A and D
A(-1,5) D(0,2)
m=(2-5)/(0+1)----> m=-3
d=√[3²+1²]----> d=√10

step 6
find the slope and the distance between point B and C
B(5,1) C(6,-2)
m=(-2-1)/(6-5)----> m=-3
d=√[3²+1²]----> d=√10
so
the slope BC=slope AD
the distance BC=distance AD

therefore

the figure is a Parallelogram

using a graph tool
see the attached figure

the answer is
the x coordinate of point D is 0