To answer this question, we need to remember that:
That is, we have to multiply by -1 each of the coordinates of the preimage to get the image.
We can check this if we identify the coordinates of the vertices of the triangle which is the preimage:
• S(-4, 2)
,
• T(-1, 3)
,
• U(-2, 1)
Therefore, to obtain the image for a rotation of 180° about the origin, we have:
• S(-4, 2) ---> (-(-4), -2) ---> S'(4, -2)
,
• T(-1, 3) ---> (-(-1), -3) ---> T'(1, -3)
,
• U(-2, 1) ---> (-(-2), -1) ---> U'(2, -1)
Hence, the coordinates of the vertices of the image are:
• S'(4, -2)
,
• T'(1, -3)
,
• U'(2, -1)
And we can see that they correspond with the ones in the image of the corresponding graph.
In summary, we can say that ∆S'T'U' is a 180° rotation about the origin of STU (second option).is the preimage:∆