The end points of one diagonal of a rhombus are (0,-8) and (8,-4). if the coordinates of the 3rd vertex are (1,0), what are the coordinates of the 4th vertex?

Diagonal of a Rhombus are perpendicular & intersects in their middle point:
Assume the diagonals intersects in H

A(0,-8), B(1,-0), C(8,-4) & D(x, y) are the vertices of the rhombus and we have to calculate D(x, y)

Consider the diagonal AC. Find the coordinate (x₁, y₁) H, the middle of AC
Coordinate  (x₁, y₁) of H, middle of A(0,-8), C(8,-4)
x₁ (0+8)/2 & y₁=(-8-4)/2 ==> H(4, -6)
Now let's calculate again the coordinate of H, middle of the diagonal BD
B(1,-0), D(x, y)
x value = (1+x)/2 & y value=(y+0)/2 ==> x= (1+x)/2 & y=y/2
(1+x)/2 & y/2 are the coordinate of the center H, already calculated, then:
H(4, -6) = [(1+x)/2 ,  y/2]==>(1+x)/2 =4 ==> x=7 & y/2 = -6 ==> y= -12

Hence the  coordinates of the 4th vertex D(7, -12)

The answer is: y would equal 3 because if you simplify the equation then it would be y= 3x+9 and if you divide 3x and 9 by 3 it would equal 3. y= 3