Question

parallelogram ABCD with vertices A(6,8) B(10,9) C(9,3) and D(5,2): 90 counterclockwise rotation about P(4,1)
A ( , )
B ( , )
C ( , )
D ( , )

Answer 1

Answer:

A(-3 , 3)  B(-4 , 7)  C(2 , 6)  D(3 , 2)

Step-by-step explanation:

Answer 2

Rotation of the parallelogram about the point P(4, 1), is done by taking

point P as the origin, and using relative coordinates for the rotation.

• $\underline{A(-3, \, 3)}$

• $\underline{B(-4, \, 7)}$

• $\underline{C(2, \, 6)}$

• $\underline{D(3, \, 2)}$

Reasons:

The given parameters are;

Vertices of the parallelogram ABCD are;

A(6, 8), B(10, 9), C(9, 3) and D(5, 2)

Point about the parallelogram is rotated = P(4, 1)

Angle of rotation = 90°

Direction of rotation = Counterclockwise

90° rotation counterclockwise of (x, y) about the origin gives → (-y, x)

Therefore, we have;

The given points relative to the point P are;

A'(6-4, 8 - 1) = A'(2, 7)

B'(10 - 4, 9 - 1) = B'(6, 8)

C'(9 - 4, 3 - 1) = C'(5, 2)

D'(5 - 4, 2 - 1) = D'(1, 1)

Therefore, we get;

A'(2, 7) $\underrightarrow {R_{180^{\circ}}}$ A''(-7, 2)

B'(6, 8) $\underrightarrow {R_{180^{\circ}}}$ B''(-8, 6)

C'(5, 2) $\underrightarrow {R_{180^{\circ}}}$ C''(-2, 5)

D'(1, 1) $\underrightarrow {R_{180^{\circ}}}$ D''(-1, 1)

Which then gives;

The image location are;

A(-7 + 4, 2 + 1) = A(-3, 3)

B''(-8 + 4, 6 + 1) = B(-4, 7)

C''(-2 + 4, 5 + 1) = C(2, 6)

D''(-1 + 4, 1 + 1) = D(3, 2)

Therefore, we have;

A(-3, 3), B(-4, 7), C(2, 6), and D(3, 2)

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