The parallelogram ABCD can be transformed onto itself by composite
transformations.
The series of transformation that would carry parallelogram ABCD onto
itself are; [tex]\underline{ \mathbf{(x + 6, \, y + 0), 180^{\circ} \, rotation, \, (x + 0, \, y + 4)}}[/tex]
Reasons:
The coordinates of the vertex of the parallelogram are;
A(-5, 1), B(-4, 3), C(-1, 3), and D(-2, 1)
- The coordinate of the point (x, y) following a rotation 180° about the origin is the point (-x, -y)
Following a rotation of the parallelogram ABCD about the origin we get;
A'(5, -1), B'(4, -3), C'(1, -3), and D'(2, -1)
Given that the parallelogram is symmetrical following rotation of 180°, we have;
The top rightmost point in the image A'B'C'D', A'(5, -1) corresponds to the top rightmost point on the preimage, point C(-1, 3)
Similarly, we have;
Point B'(4, -3), corresponds to point D(-2, 1)
Point D'(2, -1), corresponds to point B(-4, 3)
The difference in the corresponding points are;
Difference in x-values = 4 - (-2) = 6
Difference in y-values = -3 - 1 = -4
- Which gives; T₍₆, ₋₄₎, which are magnitudes similar to the third option
Therefore; by (x + 6, y + 0), 180° rotation, (x + 0, y + 4), we have;
- Adding 6 to the x-values before rotation gives;
A''(-5 + 6, 1), B''(-4 + 6, 3), C''(-1 + 6, 3), and D''(-2 + 6, 1)
A''(1, 1), B''(2, 3), C''(5, 3), and D''(4, 1)
- Rotating 180° about the origin gives;
A'''(-1, -1), B'''(-2, -3), C'''(-5, -3), and D'''(-4, -1)
- Adding 4 to the y-values gives;
A''''(-1, -1 + 4), B''''(-2, -3 + 4), C''''(-5, -3 + 4), and D''''(-4, -1 + 4)
A''''(-1, 3), B''''(-2, 1), C''''(-5, 1), and D''''(-4, 3), which are the coordinates of the image;
A''''(-1, 3) ⇔ C(-1, 3)
B''''(-2, 1) ⇔ D(-2, 1)
C''''(-5, 1) ⇔ A(-5, 1)
D''''(-4, 3) ⇔ B(-4, 3)
Therefore, the series of transformation that would carry parallelogram ABCD onto itself are; [tex]\underline{ \mathbf{(x + 6, \, y + 0), 180^{\circ} \, rotation, \, (x + 0, \, y + 4)}}[/tex]
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