)Two similar triangles are shown on the following coordinate grid:

A coordinate plane is shown. Triangle ABC has vertices A at negative 3 comma 2, B at negative 1 comma 0, and C at negative 2 comma negative 1. Triangle A prime B prime C prime has vertices A prime at 6 comma 4, B prime at 2 comma 0, and C prime at 4 comma negative 2.

Which set of transformations has been performed on triangle ABC to form triangle A′B′C′?

Comparing the points from the pre-image, ABC, to the image, A'B'C', we see that both x- and y-coordinates seem to be doubled from the pre-image to the image. However, the sign of the x-coordinates in the image is different; this means that the x-coordinates are negated. A reflection across the y-axis will negate the x-coordinates.

To double both the x- and y-coordinates, we would then apply a dilation by a factor of 2, as the scale factor gets multiplied by every coordinate.

deepakrai138 deepakrai138 · Answer 2 · 2022-01-04T00:32:25-05:00

The \(\Delta ABC\) gets reflection across Y-axis and a dilation by a scale factor of 2.

Given,

The coordinates of \(\Delta ABC\) are \(A(-3,2)\), \(B(-1,0)\) and \(C(-2,-1)\) .

The coordinates of \(\Delta A'B'C'\) are \(A'(6,4)\), \(B'(2,0)\) and \(C'(4,-2)\) .

We have to find which transformation is performed on \(\Delta ABC\) to form \(\Delta A'B'C'\).

Now take a look on the coordinates of vertex of each triangle we found that each coordinate of \(\Delta A'B'C'\) is doubled and the sign of abscissa of each coordinate of \(\Delta A'B'C'\) is changed with respect to the coordinates of \(\Delta ABC\).

Since the sign of abscissa of any coordinate of point is only changed when we reflect the point across Y-axis.

Hence the \(\Delta ABC\) gets a reflection across Y-axis.

Since the coordinates of \(\Delta A'B'C'\) gets doubled, so we can say that the \(\Delta ABC\) gets dilated by a scale factor of 2.

Thus, the \(\Delta ABC\) gets reflection across Y-axis and a dilation by a scale factor of 2.

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