Is rotating the point L(4,4) 90° ABOUT THE ORIGIN THE SAME AS REFLECTING THE POINT OVER THE X-AXIS? EXPLAIN. I get that the rule is counter clockwise and you rotate x and y and then change the value of the 1st sign.. becomes -y,x but i don't know how to answer the question.

Answer 1

No,
the point 4,4) ----> (4,-4) after a reflection in the x-a xis
but the point (4,4) ---> (-4,4) after a rotation of 90°

For a point P(x,y) the slope of PO , where O is the origin is (y-0)/(x-0) = y/x
For a point P1(-y,x) , the slope P1O = x/-y, which is the negative reciprocal of the slope of PO.
Thus they form a 90° angle as required.
Also the length of PO = √(x^2 + y^2) = length of P1O
Thus the point P has been rotated 90° about the origin.
Another example
if P is (5,1) the P1 is (-1,5). Sketch the situation to see the 90° rotation.
now look at a reflection of P in the x-axis. (5,1) ---> (5, -1)

in general for any point (a,b) you need to maintain the length so the values of a and b must be maintained.
To obtain a 90° angle the slopes have to be negative reciprocal , so the slope fraction must be flipped and the sign has to be changed
or
y/x ---> -x/y
(a,b) ---> (-b, a)

If you have studied matrices, you may recall that
cosØ -sinØ
sinØ cosØ
will rotate any point through an angle of Ø

so
cos90 -sin90
sin90 cos90
multiplies by
4 4
is
4cos90 - 4sin90
4sin90 + 4cos90
= 0-4
4 + 0
=
-4 4
or the point (-4,4)

or for my point (5,1)
5cos90 - 1sin90
5sin90 + 1cos90
=
0 -1
5 + 0
=
(-1,5)