Answer:
Step-by-step explanation:
From the graph attached,
Coordinates of the vertices of ΔOAB,
O → (0, 0)
A → (2, 4)
B → (5, 0)
If a point (h, k) is rotated 90° counterclockwise about the origin the image point will be,
(h, k) → (-k, h)
By this rule, if triangle OAB is rotated 90° counter clockwise, new vertices will follow the same rule,
O(0, 0) → O'(0, 0)
A(2, 4) → A'(-4, 2)
B(5, 0) → B'(0, 5)
By plotting these points we can get the image triangle.
