Question

triangle xyz has vertices x(0, 2), y(4, 4), and z(3, -1). triangle xyz is rotated 180° counterclockwise about z. in which quadrant is the image of point x?
a) quadrant i
b) quadrant ii
c) quadrant iii
d) quadrant iv

Explanation:

Step1: Translate the point

To rotate a point \( (x,y) \) \( 180^{\circ} \) counter - clockwise about a center \( (a,b) \), we use the formula \( (x',y')=(2a - x,2b - y) \). Here \( X(0,2) \) and \( Z(3,-1) \). So \( x = 0,y = 2,a = 3,b=-1 \).

Step2: Calculate the new coordinates

Substitute into the formula: \( x'=2\times3 - 0=6 \), \( y'=2\times(-1)-2=-2 - 2=-4 \). The image of \( X \) is \( (6,-4) \).

Answer:

D. Quadrant IV