Question

graph the image of n(-2, -2) after a rotation 90° counterclockwise around the origin.

Explanation:

Step1: Recall rotation rule

For a \(90^\circ\) counterclockwise rotation about the origin, the rule is \((x,y)\to(-y,x)\).

Step2: Apply the rule to \(N(-2,-2)\)

Here, \(x = - 2\) and \(y=-2\). Substituting into the rule: \(-y=-(-2) = 2\) and \(x=-2\). So the new coordinates are \((2,-2)\)? Wait, no, wait. Wait, the rule is \((x,y)\to(-y,x)\). So \(x=-2\), \(y = - 2\). Then \(-y=2\), \(x=-2\)? No, wait, no: \((x,y)\) becomes \((-y,x)\). So \(x=-2\), \(y=-2\), so \(-y = 2\), \(x=-2\)? Wait, no, I messed up. Wait, the correct rule for \(90^\circ\) counterclockwise rotation about the origin is \((x,y)\to(-y,x)\). So if the original point is \((x,y)\), after rotation, it's \((-y,x)\). So for \(N(-2,-2)\), \(x=-2\), \(y = - 2\). Then \(-y=2\), \(x=-2\)? Wait, no, that would be \((2,-2)\)? No, wait, no. Wait, let's take an example. If we have a point \((1,2)\), rotating \(90^\circ\) counterclockwise around origin gives \((-2,1)\). So yes, \((x,y)\to(-y,x)\). So for \((-2,-2)\), \(x=-2\), \(y=-2\). Then \(-y = 2\), \(x=-2\)? Wait, no, \(x\) in the new point is \(x\) from original? No, wait, the formula is \((x,y)\) becomes \((-y,x)\). So original \(x=-2\), \(y=-2\). So new \(x=-y = -(-2)=2\), new \(y = x=-2\). Wait, no, that's not right. Wait, no, the formula is \((x,y)\) rotated \(90^\circ\) counterclockwise is \((-y,x)\). So \(x\) coordinate of new point is \(-y\) (original \(y\) with negative sign), and \(y\) coordinate of new point is original \(x\). So for \((-2,-2)\): new \(x=-(-2)=2\), new \(y=-2\)? Wait, no, that would be \((2,-2)\), but that's a \(180^\circ\) rotation? No, wait, no, I think I mixed up. Wait, let's use the standard rotation matrix. The rotation matrix for \(90^\circ\) counterclockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). So if we have a vector \(\begin{pmatrix}x\\y\end{pmatrix}\), after rotation, it's \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}\). So yes, \((x,y)\to(-y,x)\). So for \(N(-2,-2)\), \(x=-2\), \(y=-2\). So new \(x=-y = -(-2)=2\), new \(y = x=-2\)? Wait, no, \(x\) in the matrix multiplication is the original \(x\), so new \(y\) is original \(x\). So new point is \((-y,x)= (2,-2)\)? Wait, no, that can't be. Wait, let's take a point in the third quadrant, \((-2,-2)\). Rotating \(90^\circ\) counterclockwise should take it to the second quadrant. Wait, my mistake! Oh no, I messed up the signs. Wait, the rotation matrix for \(90^\circ\) counterclockwise 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 \(x=-2\), \(y=-2\), that's \(\begin{pmatrix}-(-2)\\-2\end{pmatrix}=\begin{pmatrix}2\\-2\end{pmatrix}\)? But that's in the fourth quadrant. Wait, that's wrong. Wait, no, wait, if the point is \((-2,-2)\) (third quadrant), rotating \(90^\circ\) counterclockwise around origin: the direction is counterclockwise, so from third quadrant, rotating \(90^\circ\) counterclockwise would go to the second quadrant. So my formula must be wrong. Wait, no, wait, maybe I mixed up clockwise and counterclockwise. Wait, the rotation matrix for \(90^\circ\) clockwise is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), which gives \((y,-x)\). And \(90^\circ\) counterclockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), which gives \((-y,x)\). Wait, let's take a point \((1,0)\), rotating \(90^\circ\) counterclockwise should be \((0,1)\). Using the formula: \((-0,1)=(0,1)\), correct. Point \((0,1)\), rotating \(90^\circ\) counterclockwise should be \((-1,0)\). Using formula: \((-1,0)\), correct. Point \((-1,0)\), rotating \(90^\circ\) counterclockwise should be \((0,-1)\). Formula: \((-0,-1)=(0,-1)\), correct. Point \((0,-1)\), rotating \(90^\circ\) counterclockwise should be \((1,0)\). Formula: \((-(-1),0)=(1,0)\), correct. Now take a point in third quadrant: \((-2,-2)\). Using formula: \((-(-2),-2)=(2,-2)\). Wait, that's fourth quadrant. But that seems wrong. Wait, no, let's plot it. Original point: \((-2,-2)\) (3rd quadrant). Rotating \(90^\circ\) counterclockwise: the x and y coordinates swap, and the new x is the negative of the original y. Wait, original x=-2, y=-2. Swap x and y: (-2,-2) becomes (-2,-2) swapped? No, wait, no, the rule is (x,y) → (-y, x). So x=-2, y=-2: -y=2, x=-2. So (2, -2)? Wait, that's (2,-2), which is 4th quadrant. But when you rotate ( -2,-2 ) 90 degrees counterclockwise around origin, let's think about the angle. The original point is at 225 degrees from positive x-axis (since it's in third quadrant, 180 + 45 = 225). Rotating 90 degrees counterclockwise adds 90 degrees, so 225 + 90 = 315 degrees, which is in the fourth quadrant (315 degrees is equivalent to -45 degrees, so (r cos 315, r sin 315) = (r*(√2/2), r*(-√2/2))). The original point has r = √[(-2)^2 + (-2)^2] = √8 = 2√2. So new coordinates: (2√2 cos 315, 2√2 sin 315) = (2√2*(√2/2), 2√2*(-√2/2)) = (2, -2). Oh! So my initial intuition was wrong. Rotating ( -2,-2 ) 90 degrees counterclockwise around origin does give (2, -2). Because 225 + 90 = 315 degrees, which is in the fourth quadrant. So the formula is correct. So the image of N(-2,-2) after 90 degrees counterclockwise rotation around origin is (2, -2).

Answer:

The image of \(N(-2,-2)\) after a \(90^\circ\) counterclockwise rotation around the origin is the point with coordinates \((2, - 2)\). To graph it, locate the point where \(x = 2\) and \(y=-2\) on the coordinate plane.