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 point \(X\) relative to \(Z\)

To rotate a point \(P(x,y)\) \(180^{\circ}\) about a center \(C(a,b)\), we first find the vector from \(C\) to \(P\). The vector from \(Z(3, - 1)\) to \(X(0,2)\) is \((0 - 3,2-(-1))=(-3,3)\).

Step2: Apply \(180^{\circ}\) rotation (reverse the vector)

A \(180^{\circ}\) counter - clockwise rotation about a point reverses the vector. So the new vector is \((3,-3)\).

Step3: Translate back to get the image of \(X\)

Add the new vector to the center \(Z\). The coordinates of the image of \(X\) are \((3 + 3,-1+(-3))=(6,-4)\)

Answer:

D. Quadrant IV