We'll label the positive #x#-axis east (to the right), and the positive #y#-axis north (upward).
To find the magnitude of this displacement, we can use the formula
#r = sqrt(x^2 + y^2)#
where #r# is the magnitude of the displacement vector,
#x# is how far along the #x#-axis it was displaced, (how far east or west it is) and
#y# is how far along the #y#-axis it was displaced (how far north or south it is).
This equation is essentially the distance formula, but with the starting position at the origin.
It traveled #2.5 "m"# east, so we'll substitute #2.5 "m"# for the #x# value, and it traveled #6.0 "m"# north, so we'll substitute #6.0 "m"# for the #y# value. This gives the magnitude of the displacement
#r = sqrt((2.5"m")^2 + (6.0"m")^2) = color(red)(6.5 "m"#
The direction of the displacement vector is found according to the equation
#tantheta = y/x#
And so the angle, #theta#, of the displacement vector (with respect to the positive #x#-axis) is
#theta = arctan(y/x) = arctan((6.0"m")/(2.5"m")) = color(blue)(67^o#
(The function #arctan# is the same as the function #tan^-1#, which you're more likely to see on a calculator.)
The displacement vector of this motion is thus #6.5 "m"#, #67^o# north of east (counterclockwise from the positive #x#-axis, which is most often where angles are measured from).
