Question

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

Explanation:

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:

\(x' = 01 + (-1)0 = 0\)

\(y' = 1*1 + 0*0 = 1\)

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: \((1, -7)\)