Question

triangle abc has coordinates a(6,3), b(8,7), and c(8,2), and after a rotation, its image has coordinates a(6,3), b(4,−1), and c(4,4). which statement describes the rotation?
a) rotation of 90° counterclockwise about point a
b) rotation of 180° clockwise about point a
c) rotation of 180° counterclockwise about the origin
d) rotation of 270° counterclockwise about point a

Explanation:

Step1: Analyze Point A

Since \( A(6,3) \) and \( A'(6,3) \) are the same, the center of rotation must be point \( A \) (because the center of rotation remains fixed). So we can eliminate option C (which is about the origin).

Step2: Analyze Vector from A to B

First, find the vector \( \overrightarrow{AB} \). \( A(6,3) \), \( B(8,7) \), so \( \overrightarrow{AB}=(8 - 6,7 - 3)=(2,4) \). After rotation, \( B'(4,-1) \), so the vector \( \overrightarrow{AB'}=(4 - 6,-1 - 3)=(-2,-4) \). Notice that \( (-2,-4)=-(2,4) \), which is a \( 180^\circ \) rotation (since a \( 180^\circ \) rotation about a point \( (x_0,y_0) \) transforms a vector \( (x,y) \) to \( (-x,-y) \) relative to the center). Let's check the direction: clockwise or counterclockwise? A \( 180^\circ \) rotation is the same clockwise or counterclockwise in terms of the vector transformation (since \( 180^\circ \) is symmetric). Let's also check point C. \( C(8,2) \), vector \( \overrightarrow{AC}=(8 - 6,2 - 3)=(2,-1) \). After rotation, \( C'(4,4) \)? Wait, wait, the problem says \( C'(4,4) \)? Wait, the original problem says \( C'(4,4) \)? Wait, the user's problem says \( C'(4,4) \)? Let me re - check. The user's problem: "A'(6,3), B'(4, - 1), and C'(4,4)". Wait, \( \overrightarrow{AC}=(8 - 6,2 - 3)=(2,-1) \), \( \overrightarrow{AC'}=(4 - 6,4 - 3)=(-2,1)=-(2,-1) \), which is also a \( 180^\circ \) rotation (since \( (x,y)\to(-x,-y) \) relative to center \( A \)). A \( 90^\circ \) counterclockwise rotation about \( A \) would transform \( (2,4) \) to \( (-4,2) \) (using the rotation formula \( (x,y)\to(-y,x) \) for \( 90^\circ \) counterclockwise about a point \( (x_0,y_0) \): new \( x = x_0-(y - y_0) \), new \( y = y_0+(x - x_0) \)). Let's test: for \( B(8,7) \), \( x=8,y = 7,x_0 = 6,y_0 = 3 \). \( 90^\circ \) counterclockwise: new \( x=6-(7 - 3)=6 - 4 = 2 \), new \( y=3+(8 - 6)=3 + 2 = 5 \), which is not \( (4,-1) \). A \( 270^\circ \) counterclockwise rotation is equivalent to a \( 90^\circ \) clockwise rotation, which would transform \( (2,4) \) to \( (4,-2) \) (using \( (x,y)\to(y,-x) \) for \( 90^\circ \) clockwise about center). Let's test: \( x = 8,y = 7,x_0=6,y_0 = 3 \). New \( x=3+(7 - 3)=7 \)? Wait, maybe my formula is off. But the key is that the vector from \( A \) to \( B \) and \( A \) to \( B' \) is a scalar multiple of \( - 1 \), indicating \( 180^\circ \) rotation. And since the center is \( A \) (because \( A \) is fixed), and the rotation is \( 180^\circ \) (either clockwise or counterclockwise, but option B says "rotation of \( 180^\circ \) clockwise about point A" and option D is \( 270^\circ \) counterclockwise about A, option A is \( 90^\circ \) counterclockwise about A). Since the vector transformation is \( (x,y)\to(-x,-y) \) relative to A, which is a \( 180^\circ \) rotation. So the correct option is B (since a \( 180^\circ \) rotation clockwise about A is the same as counterclockwise in terms of the result, and our vector analysis shows a \( 180^\circ \) rotation about A).

Answer:

B. rotation of \( 180^\circ \) clockwise about point A