Answer:
The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)
The coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)
Step-by-step explanation:
Square JKLM with vertices
We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.
In other words, the rule to rotate a point by 90° counterclockwise.
P(x, y) → P'(-y, x)
As we are given that J(1, -3), so the coordinates of J' will be:
J(1, -3) → J'(3, 1)
Therefore, the coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1).
When the point is rotated by 90° clockwise, we need to switch x and y, make x negative.
In other words, the rule to rotate a point by 90° clockwise.
P(x, y) → P'(y, -x)
As we are given that J(1, -3), so the coordinates of J' will be:
J(1, -3) → J'(-3, -1)
Therefore, the coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)
(10)(13)