Question

A polygon has coordinates A(-7, 8), B(-4, 6), C(-4, 3), D(-8, 3), and E(-9, 6). What are the coordinates of its image, polygon A′B′C′D′E′, after a 270° counterclockwise rotation about the origin and a translation 2 units to the left and 3 units up? A′(-6, 10), B′(-4, 7), C′(-1, 7), D′(-1, 11), E′(-4, 12) A′(6, 10), B′(4, 7), C′(1, 7), D′(1, 11), E′(4, 12) A′(5, 11), B′(3, 8), C′(0, 8), D′(0, 12), E′(3, 13) A′(-5, 11), B′(-3, 8), C′(0, 8), D′(0, 12), E′(-3, 13)

Answer 1

The correct answer is third set of points: A′(6, 10), B′(4, 7), C′(1, 7), D′(1, 11), E′(4, 12) 

We can figure out this answer, just by looking at the first point A. If we rotate it 270 degree counterclockwise, if will go from the second quadrant to the first quadrant. 

The new point will be (8, 7). If we move left 2 and up 3, we will be at the point (6, 10). Only the second choice matches this.

Answer 2

Answer:  the correct option is

(B) A′(6, 10), B′(4, 7), C′(1, 7), D′(1, 11), E′(4, 12)

Step-by-step explanation:  Given that a polygon has coordinates A(-7, 8), B(-4, 6), C(-4, 3), D(-8, 3), and E(-9, 6). ABCDE is rotated counterclockwise by 270 degrees and translated 2 units to the left and 3 units up to form the image A'B'C'D'E'.

We are to find the co-ordinates of the image A'B'C'D'E'.

If a point (x, y) is rotated 270 degrees counterclockwise and then translated 2 units to the left and 3 units up, then its co-ordinates changes as follows

$(x,y)~~~\Rightarrow~~~(y-2,-x+3).$

So, the co-ordinates of the image A'B'C'D'E' are as follows :

$A(-7,8)~~~\Rightarrow~~~A'(8-2,7+3)=A'(6,10),\\\\B(-4,6)~~~\Rightarrow~~~B'(6-2,4+3)=B'(4,7),\\\\C(-4,3)~~~\Rightarrow~~~C'(3-2,4+3)=C'(1,7),\\\\D(-8,3)~~~\Rightarrow~~~D'(3-2,8+3)=D'(1,11),\\\\E(-9,6)~~~\Rightarrow~~~E'(6-2,9+3)=E'(4,12).$

Thus, the required co-ordinates of the image A'B'C'D'E' are A′(6, 10), B′(4, 7), C′(1, 7), D′(1, 11) and E′(4, 12).

Option (B) is CORRECT.