Which red triangle shows a 90° counterclockwise rotation of the blue triangle? Check all that apply. On a coordinate plane, a blue triangle has points (negative 1, 4), (negative 5, 4), (negative 1, 1). A red triangle has points (negative 4, negative 1), (negative 1, negative 1), (negative 4, negative 5). On a coordinate plane, a blue triangle has points (1, 1), (4, 5), (4, 1). A red triangle has points (negative 1, 1), (negative 1, 4), (negative 5, 4). On a coordinate plane, a blue triangle has points (0, 1), (4, 4), (4, 1). A red triangle has points (negative 4, 1), (negative 4, 4), (0, 1). On a coordinate plane, a blue triangle has points (negative 5, 4), (negative 1, 4), (negative 1), 1). A red triangle has points (1, 1), (4, 1), (4, 5). On a coordinate plane, a blue triangle has points (negative 2, 5), (4, 5), (4, 1). A red triangle has points (negative 1, 4), (negative 5, 4), (negative 5, negative 2).

Question:

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

The correctly rotated red triangles are those of (1), (2), and (5)

Step-by-step explanation:

In a 90° counterclockwise rotation, every x and y points of the original triangle are switched while the y is turned negative, as shown in the following equation;

(x, y) to (-y, x)

Therefore, the triangles that undergo a 90° counterclockwise rotation are as follows;

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

(-1, 4) → (-4, -1)

(-5, 4) → (-4, -5)

(-1, 1) → (-1, -1)

Correctly rotated 90° counterclockwise

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

(1, 1) → (-1, 1)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

(0, 1) → (0, 1) ≠ (-1, 0)

(4, 4) → (-4, 4)

(4, 1) → (-4, 1)

Not correctly rotated 90° counterclockwise

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

(-5, 4) → (4, 5) ≠ (-4, -5)

(-1, 4) → (4, 1)

(-1, 1) → (1, 1)

Not correctly rotated 90° counterclockwise

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

(-2, 5) → (-5, -2)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise.

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)