Find the magnitude & direction of the vector jk given points . j(-6,-4) & k(-10,-4)

There is a formula about the distance between two points in the plane.
M(a,b) and Т(с,d)
The distance between M and N is L=sqrt[(a-c)^2+(b-d)^2)]
So the magnitude of this vector is  sqrt[(-6+10)^2+(-4+4)^2]=sqrt(16+0)=4
b=d This means that vector JK is parallel to the OX axis(there is a picture)

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