Image of point !1,1) under counterclockwise rotation will be located at (--root3/2, 1/2)

the distance d = ax' + by' + c)/ Sq root a^2+b^2

d = (1 .1 + 3 . 1 + 4)/ Sqroot 10

d = 8/sqrt 10

It can be easily seen if we graph it with the unit circle.

Rotating the point (1,0) by π/3 will lead us to a point on the unit circle where theta = π/3.

x = cos theta

= cos π/3

= 1/2

y = sin theta

= sin π/3

= sqrt3/2

Thus, the image we need to find is (1/2, sqrt3/2)