Solving for Electric Field at an Angle: Ex = 10.0V/m

[TTauri]

Homework Statement

In a certain region of space, there is a uniform electric field of magnitude
10.0 V m−1 directed at 45.0◦ to the positive direction of both the x and y axes. (There is no z
component of the electric field.) The electric potential at the origin is +100 V. What is the electric
potential at the point where x = 4.00 m and y = 3.00 m?

Relevant Equations

Ex = -dV/dx

Ex =10.0Vm-1
dx= d²=x²+y²
x=4.00m
y=3.00m
d²=4.00²+3.00²
d=5.00m
Ex = -dV/dx
10.0Vm^-1=-dV/5.00m
10.0Vm^-1*5.00m=-dV
-50.0V=dV
So from origin at 100V-50.0V = 50.0V

But the solution I am given gives the answer at 50.5V and the information "directed at 45.0◦ " does not seem to have been used so may related to the difference.
I have hunted a bit round the internet and through textbooks but cannot find this formulation with the field being directed at an angle.

 

[Doc Al]

Realize that direction matters. If you drew a line from the origin to the point in question, what angle would the field make with that line?