the figure below has a point marked with a large dot. first, rotate the figure 270° clockwise about the origin. then, give the coordinates of the marked point in the original figure and the final fig

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# Answer:
Original point: $(1, -7)$
Final point after $270^\circ$ clockwise rotation: $(-7, -1)$

(Note: Assuming the marked dot in the original figure is at $(1, -7)$ (from the grid analysis, as the dot is at $x=1$, $y=-7$ approximately). The rotation rule for $270^\circ$ clockwise (or $90^\circ$ counterclockwise) is $(x, y) \to (-y, x)$. So applying to $(1, -7)$: $x = 1$, $y = -7$, new $x = -(-7)=7$? Wait, correction: The correct rule for $270^\circ$ clockwise rotation about origin is $(x, y) \mapsto (y, -x)$? Wait no, let's recall:

- $90^\circ$ clockwise: $(x, y) \to (y, -x)$
- $180^\circ$ clockwise: $(x, y) \to (-x, -y)$
- $270^\circ$ clockwise: $(x, y) \to (-y, x)$

Wait, let's take a point $(a, b)$. Rotating $270^\circ$ clockwise is equivalent to rotating $90^\circ$ counterclockwise. The rule for $90^\circ$ counterclockwise is $(x, y) \to (-y, x)$. So if original point is $(1, -7)$, then:

$x = 1$, $y = -7$

After $270^\circ$ clockwise (same as $90^\circ$ counterclockwise):

New $x = -y = -(-7) = 7$

New $y = x = 1$

Wait, I must have misread the original point. Looking at the graph, the marked dot (blue) is at $x=1$, $y=-7$? Wait no, the lower figure: the dot is at (1, -7)? Wait the grid: x-axis from -8 to 8, y-axis from -8 to 8. The lower figure (the triangle) has a dot at (1, -7)? Wait no, let's check coordinates. Let's assume the original marked point (before rotation) is $(1, -7)$. Wait, no, maybe the original figure is the one with the dot at, say, (1, -7), and after rotating 270 degrees clockwise, the coordinates change. Wait, perhaps the original point is $(1, -7)$, and after $270^\circ$ clockwise rotation, the new coordinates are $(-7, -1)$? No, let's rederive the rotation rule.

Wait, the standard rotation matrices:

- $90^\circ$ clockwise: $\begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}$
- $180^\circ$ clockwise: $\begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix}$
- $270^\circ$ clockwise: $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$

So for a point $(x, y)$, applying $270^\circ$ clockwise rotation:

$x' = 0 \cdot x + (-1) \cdot y = -y$

$y' = 1 \cdot x + 0 \cdot y = x$

So $(x, y) \to (-y, x)$

So if original point is $(1, -7)$, then:

$x' = -(-7) = 7$

$y' = 1$

So $(7, 1)$? But that doesn't match. Wait, maybe the original point is $(-7, 1)$? No, the graph: the left figure (the triangle) has a dot at (-7, -1)? Wait, the left triangle: the dot is at x=-7, y=-1? Let's check the grid. The x-axis: from left, -8, -7, -6, etc. The y-axis: -1, -2, etc. So the left triangle's dot is at (-7, -1). Then, rotating that point $270^\circ$ clockwise about origin.

Original point: $(-7, -1)$

Applying rotation rule $(x, y) \to (-y, x)$:

$x' = -(-1) = 1$

$y' = -7$

So final point: $(1, -7)$. But that's the reverse. Wait, maybe the original figure is the lower triangle with dot at (1, -7), and after rotating 270 degrees clockwise, it moves to (-7, -1)? No, I think I messed up the rotation direction.

Wait, $270^\circ$ clockwise is the same as $90^\circ$ counterclockwise. Let's take a simple point, say (1, 0). Rotating $270^\circ$ clockwise: it should go to (0, -1)? Wait no, (1,0) rotated 90 degrees clockwise is (0, -1), 180 is (-1, 0), 270 is (0, 1). Wait, no:

- Rotating (1,0) 90° clockwise: (0, -1)
- 180° clockwise: (-1, 0)
- 270° clockwise: (0, 1)

Using the matrix for 270° clockwise: $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$

So (1,0) multiplied by this matrix:

$x' = 0*1 + (-1)*0 = 0$

$y' = 1*1 + 0*0 = 1$

So (0,1), which is correct (270° clockwise from (1,0) is (0,1)).

Another example: (0,1) rotated 270° clockwise:

$x' = -1$, $y' = 0$ → (-1, 0)

Which is correct.

So the rule is $(x, y) \to (-y, x)$ for 270° clockwise.

Now, looking at the graph: the marked dot (blue) is in the lower triangle, at (1, -7)? Wait, no, the lower triangle is at x=1, y=-5 to -7? Wait, the grid lines: each square is 1 unit. The lower triangle has a dot at (1, -7)? Let's assume the original point is (1, -7). Then rotating 270° clockwise:

$x' = -(-7) = 7$

$y' = 1$

So (7, 1). But that's not matching. Alternatively, maybe the original point is (-7, -1), as in the left triangle. Let's check:

Original point: (-7, -1)

Rotate 270° clockwise:

$x' = -(-1) = 1$

$y' = -7$

So (1, -7), which is the lower triangle's dot. Ah! So the original figure is the left triangle with dot at (-7, -1), and after rotating 270° clockwise about origin, the dot moves to (1, -7). Wait, but the problem says "First, rotate the figure 270° clockwise about the origin. Then, give the coordinates of the marked point in the original figure and the final figure."

So original figure: left triangle, marked point at (-7, -1).

After rotating the figure 270° clockwise about origin, the marked point's coordinates are (1, -7).

Wait, but let's confirm with the rotation rule. Original point: (-7, -1).

Applying $270^\circ$ clockwise rotation: $(x, y) \to (-y, x)$

So $x' = -(-1) = 1$

$y' = -7$

Yes, so (1, -7).

So original coordinates: $(-7, -1)$

Final coordinates: $(1, -7)$

Wait, but maybe I had the original figure wrong. Let's re-express:

The left triangle (with the dot) is the original figure. Rotating it 270° clockwise about origin. The marked point in the original figure is at (-7, -1). After rotation, using the rule $(x, y) \to (-y, x)$, we get (1, -7), which is the dot in the lower triangle.

Yes, that makes sense. So:

Original point: $(-7, -1)$

Final point after $270^\circ$ clockwise rotation: $(1, -7)$

Wait, but earlier matrix gave (0,1) for (1,0) rotated 270° clockwise, which is correct. So applying to (-7, -1):

$x' = -y = -(-1) = 1$

$y' = x = -7$

Yes, so (1, -7).

So the correct coordinates are:

Original: $(-7, -1)$

Final: $(1, -7)$

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