Question

what is the image of the point (-3,5) after a rotation of 90° counterclockwise about the origin? answer attempt 1 out of 2 (□,□) submit answer

Explanation:

Step1: Recall 90° CCW rotation rule

For a point \((x, y)\), rotating 90° counterclockwise about the origin transforms it to \((-y, x)\).

Step2: Apply the rule to \((-3, 5)\)

Here, \(x = -3\) and \(y = 5\). Substitute into the rule: \(-y = -5\), \(x = -3\)? Wait, no—wait, the rule is \((x, y) \to (-y, x)\). So \(x=-3\), \(y = 5\): new \(x\) is \(-y=-5\), new \(y\) is \(x = -3\)? Wait, no, wait: Wait, 90° counterclockwise rotation formula is \((x, y) \mapsto (-y, x)\). Let's check: original point \((x,y)=(-3,5)\). So \(-y = -5\), \(x = -3\)? Wait, no, wait, maybe I mixed up. Wait, 90° counterclockwise: the formula is \((x, y) \to (-y, x)\). So for \((-3,5)\), \(x=-3\), \(y = 5\). So new \(x\) is \(-y = -5\), new \(y\) is \(x = -3\)? Wait, no, that can't be. Wait, let's take a simple point, like (1,0). Rotating 90° CCW around origin should be (0,1)? Wait no, (1,0) rotated 90° CCW is (0,1)? Wait no, 90° CCW rotation: the standard formula is \((x, y) \to (-y, x)\). Wait, (1,0): \(x=1\), \(y=0\). Then \(-y = 0\), \(x=1\)? No, that's (0,1)? Wait, no, (1,0) rotated 90° CCW is (0,1)? Wait, no, actually, 90° counterclockwise rotation: the coordinates \((x,y)\) become \((-y, x)\). Wait, (1,0): \(-y = 0\), \(x = 1\)? No, that's (0,1)? Wait, no, (1,0) rotated 90° CCW is (0,1)? Wait, no, let's use the rotation matrix. The rotation matrix for 90° CCW is \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\). So multiplying by \(\begin{pmatrix}x \\ y\end{pmatrix}\) gives \(\begin{pmatrix}-y \\ x\end{pmatrix}\). So for (1,0): \(\begin{pmatrix}-0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 1\end{pmatrix}\), which is correct (rotating (1,0) 90° CCW around origin is (0,1)). Now, take (0,1): rotating 90° CCW gives (-1, 0), which matches the matrix: \(\begin{pmatrix}-1 \\ 0\end{pmatrix}\). So the formula is correct: \((x,y) \to (-y, x)\). Now, apply to (-3,5): \(x=-3\), \(y=5\). So new \(x\) is \(-y = -5\), new \(y\) is \(x = -3\)? Wait, no, wait: \(x=-3\), \(y=5\). So \(-y = -5\), \(x = -3\)? So the new point is (-5, -3)? Wait, no, wait, let's check with (2,3). Rotating 90° CCW should be (-3,2). Let's apply the formula: \(x=2\), \(y=3\). \(-y=-3\), \(x=2\). So (-3,2), which is correct. So for (-3,5): \(x=-3\), \(y=5\). So \(-y = -5\), \(x = -3\)? Wait, no, \(x\) in the original is -3, so new \(y\) is \(x=-3\), new \(x\) is \(-y=-5\). So the new point is (-5, -3)? Wait, no, wait, (2,3) becomes (-3,2). So (x,y) → (-y, x). So (2,3) → (-3,2). So yes, x becomes -y, y becomes x. So for (-3,5): x=-3, y=5. So new x: -y = -5, new y: x = -3. So the image is (-5, -3)? Wait, no, wait, let's plot (-3,5). Rotating 90° counterclockwise: from quadrant II, rotating 90° CCW would move to quadrant III? Wait, (-3,5) is in QII. Rotating 90° CCW: the x and y swap, and x becomes negative of original y. Wait, maybe I had the formula wrong. Wait, 90° clockwise is (y, -x), 90° counterclockwise is (-y, x). Let's confirm with (0,1): 90° CCW should be (-1, 0). Using the formula: (0,1) → (-1, 0). Yes, because -y = -1, x=0? Wait no, (0,1): x=0, y=1. So -y = -1, x=0? No, that gives (-1, 0), which is correct. (1,0) → (0,1)? Wait no, (1,0) rotated 90° CCW is (0,1)? Wait, no, (1,0) is on the x-axis. Rotating 90° CCW around origin: it should go to (0,1), which is on the y-axis. Using the formula: (1,0) → (-0, 1) = (0,1), correct. (0,1) → (-1, 0), correct. (1,1) → (-1, 1), correct? Wait, (1,1) rotated 90° CCW: the vector (1,1) rotated 90° CCW is (-1,1), yes. So formula holds. So back to (-3,5): x=-3, y=5. So -y = -5, x = -3. So the new point is (-5, -3)? Wait, no, wait: x=-3, y=5. So -y is -5, x is -3. So (x,y) → (-y, x) → (-5, -3). Wait, but let's check with another approach. The rotation matrix for 90° CCW is \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\). So multiplying by the vector \(\begin{pmatrix}x \\ y\end{pmatrix}\) gives \(\begin{pmatrix}-y \\ x\end{pmatrix}\). So for \(\begin{pmatrix}-3 \\ 5\end{pmatrix}\), the product is \(\begin{pmatrix}-5 \\ -3\end{pmatrix}\). Yes, that's correct. So the image is (-5, -3).

Answer:

\((-5, -3)\)