We are told that the coordinate grid below shows the final image of a point that was rotated $9 0^{\circ}$ clockwise about the origin, dilated by a scale factor of $2,$ and shifted $7$ units right.

To determine the original coordinates of the point, let's reverse each of these steps starting with the last one. The point was shifted $7$ units right, so we need to shift it horizontally $7$ units left. Let's do this!

Before that, the point was dilated by a scale factor of

2 .

Let's recall that if a point

(x, y)

is dilated about the origin, both

x -

and

y -

coordinate of the point are multiplied by the scale factor

k

of the dilation.

(x, y) \to (k x, k y)

In our case the scale of the dilation is

2,

so the original coordinates were doubled. From the diagram, we know that the coordinates of the image are

(- 6, 2) .

Dividing each coordinate by

2,

we can find the coordinates of

P

before dilation.

(\frac{- 6}{2}, \frac{2}{2}) = (- 3, 1)

Let's plot it on the coordinate plane!

Before dilation, the point was rotated $9 0^{\circ}$ clockwise around the origin. Hence, to find the initial position of the point, let's rotate it $9 0^{\circ}$ counterclockwise.

We conclude that the original coordinates of the point $P$ are $(- 1, - 3) .$

Exercise