The coordinates of the vertices of the image of the trapezoid are given as;
A'(x, y) = (- 4, 1), B'(x, y) = (- 2, 1), C'(x, y) = (- 5 / 2, 5 / 2) , D'(x, y) = (- 7 / 2, 5 / 2).
<h3>How to find the image of a trapezoid by dilation?</h3>
In this question we have a representation of a trapezoid, whose image has to be generated by a kind of rigid transformation known as dilation, whose equation is described :
P'(x, y) = O(x, y) + k · [P(x, y) - O(x, y)]
Where O(x, y) - Center of dilation
k - Scale factor
And P(x, y) - Coordinates of the original point, P'(x, y) - Coordinates of the resulting point.
Since k = 1 / 2, A(x, y) = (- 5, - 2), B(x, y) = (- 1, - 2), C(x, y) = (- 2, 1), D(x, y) = (- 4, 1), O(x, y) = (- 3, 4),
Therefore, the coordinates of the vertices of the image are:
Point A'
A'(x, y) = O(x, y) + k · [A(x, y) - O(x, y)]
A'(x, y) = (- 3, 4) + (1 / 2) [(- 5, - 2) - (- 3, 4)]
A'(x, y) = (- 3, 4) + (1 / 2) (- 2, - 6)
A'(x, y) = (- 3, 4) + (- 1, - 3)
A'(x, y) = (- 4, 1)
Point B';
B'(x, y) = O(x, y) + k [B(x, y) - O(x, y)]
B'(x, y) = (- 3, 4) + (1 / 2) [(- 1, - 2) - (- 3, 4)]
B'(x, y) = (- 3, 4) + (1 / 2) (2, - 6)
B'(x, y) = (- 3, 4) + (1, - 3)
B'(x, y) = (- 2, 1)
Point C';
C'(x, y) = O(x, y) + k · [C(x, y) - O(x, y)]
C'(x, y) = (- 3, 4) + (1 / 2) [(- 2, 1) - (- 3, 4)]
C'(x, y) = (- 3, 4) + (1 / 2) (1, - 3)
C'(x, y) = (- 3, 4) + (1 / 2, - 3 / 2)
C'(x, y) = (- 5 / 2, 5 / 2)
Point D'
D'(x, y) = O(x, y) + k [D(x, y) - O(x, y)]
D'(x, y) = (- 3, 4) + (1 / 2) [(- 4, 1) - (- 3, 4)]
D'(x, y) = (- 3, 4) + (1 / 2) (- 1, - 3)
D'(x, y) = (- 3, 4) + (- 1 / 2, - 3 / 2)
D'(x, y) = (- 7 / 2, 5 / 2)
To learn more on dilations:
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= -2