Line segment su is dilated to create s'u' using the dilation rule dq,2.5. what is the distance, x, between points u' and u? .

Assuming the centre of dilation Q is the origin(0,0)
If preimage U(x1,y1), then image U'(2.5x1, 2.5y1).
The horizontal distance between U' and U would be
dx=2.5x1-x1 = 1.5x1,
the corresponding vertical distance is
dy=2.5y1-y1 = 1.5y1
The oblique distance between U and U' is therefore
x=sqrt(dx^2+dy^2)

Example: if U(2,3) is dilated about (0,0) with scale factor 2.5,
U'(2.5*2, 2.5*3)=U'(5,7)
Horizontal distance = (5-2)=3
Vertical distance = (7.5-3)=4.5
Oblique distance = sqrt(3^2+4.5^2)=sqrt(9+20.25)=sqrt(29.25)=5.41 approx.

If Q is NOT the origin, but Q(x0,y0)
then
U(x1,y1)
U'(2.5(x1-x0)+x0,2.5(y1-y0)+y0) = U'(2.5x1-1.5x0, 2.5y1-1.5y0)
The horizontal & vertical distances between U and U' is therefore
dx=2.5x1-1.5x0-x1=1.5(x1-x0)
dy=2.5y1-1.5y0-y1=1.5(y1-y0)
The oblique distance between U and U' is therefore
x=sqrt(dx^2+dy^2)
=sqrt(1.5^2(x-x0)^2+1.5^2(y-y0)^2
=1.5sqrt((x-x0)^2+(y-y0)^2)

Example: U(5,5) dilated about (2,1) with scale factor of 2.5

U'(2.5(5-2)+2,2.5(5-1)+1=U'(2.5(3)+2,2.5(4)+1)=U'(9.5,11)
horizontal distance, dx=9.5-5=4.5
vertical distance, dy = 11-5 =  6
oblique distance
= sqrt(dx^2+dy^2)
= sqrt(4.5^2+6^2)
= sqrt(20.25+36)
= sqrt(56.25)
= 7.5