Question

enter the coordinates of the image after △jkl is rotated 90° counterclockwise around the origin.
table with columns jkl and jkl, rows: j(-2,4) with ?, k(-2,2) with ?, l(-4,2) with ?
graph of △jkl on a coordinate grid, showing points j, k, l and a rotated triangle below

Explanation:

Step1: Recall rotation rule

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

Step2: Rotate \(J(-2,4)\)

Using the rule, \(x = - 2\), \(y = 4\). So the new coordinates are \((-4,-2)\)? Wait, no, wait: \((x,y)\to(-y,x)\). So for \(J(-2,4)\), \(-y=-4\), \(x = - 2\)? Wait, no, I messed up. Wait, the correct rule is: when rotating \(90^\circ\) counterclockwise about the origin, the transformation is \((x,y)\mapsto(-y,x)\). Wait, let's check again. Let's take a point \((a,b)\). After \(90^\circ\) counterclockwise rotation, the new point \((x',y')\) is given by \(x'=-b\), \(y' = a\). So for \(J(-2,4)\): \(a=-2\), \(b = 4\). So \(x'=-4\), \(y'=-2\)? Wait, no, wait, no. Wait, let's take an example. If we have a point \((1,0)\), rotating \(90^\circ\) counterclockwise gives \((0,1)\). Using the formula \((x,y)\to(-y,x)\): \((1,0)\to(0,1)\), which is correct. Another example: \((0,1)\) rotated \(90^\circ\) counterclockwise is \((-1,0)\). Using the formula: \((0,1)\to(-1,0)\), correct. So for \((x,y)\), the new coordinates are \((-y,x)\). So for \(J(-2,4)\): \(x=-2\), \(y = 4\). So \(-y=-4\), \(x=-2\)? Wait, no, \(x\) in the original is \(-2\), so the new \(y'\) is \(x=-2\), and new \(x'\) is \(-y=-4\). So \(J'(-4,-2)\)? Wait, no, wait, no. Wait, let's do it step by step. Let's take \(J(-2,4)\). The original coordinates: \(x=-2\), \(y = 4\). After \(90^\circ\) counterclockwise rotation, the new \(x\)-coordinate is \(-y\) and new \(y\)-coordinate is \(x\). So \(x'=-y=-4\), \(y'=x=-2\). So \(J'(-4,-2)\)? Wait, but looking at the graph, the lower triangle: let's check \(K(-2,2)\). For \(K(-2,2)\), \(x=-2\), \(y = 2\). So \(x'=-y=-2\), \(y'=x=-2\). So \(K'(-2,-2)\)? Wait, no, the lower triangle has a point at \((-2,-4)\)? Wait, maybe I made a mistake. Wait, no, let's re-express the rule. The correct rule for \(90^\circ\) counterclockwise rotation about the origin is \((x,y)\to(-y,x)\). Wait, let's take \(J(-2,4)\): \(x=-2\), \(y = 4\). So \(-y=-4\), \(x=-2\). So \(J'(-4,-2)\)? But in the lower triangle, the point corresponding to \(J\) is at \((-4,-4)\)? Wait, no, the graph shows the lower triangle with a point at \((-2,-4)\)? Wait, maybe I got the rotation direction wrong. Wait, the problem says counterclockwise. Wait, maybe the rule is \((x,y)\to(y,-x)\) for clockwise? No, no. Let's confirm the rotation rule.

The standard 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 becomes \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}\). So the new coordinates are \((-y,x)\). So for \(J(-2,4)\): \(x=-2\), \(y = 4\). So \(-y=-4\), \(x=-2\). So \(J'(-4,-2)\)? Wait, but in the graph, the lower triangle has a point at \((-4,-4)\)? No, looking at the graph, the upper triangle has \(J\) at \((-2,4)\), \(K\) at \((-2,2)\), \(L\) at \((-4,2)\). The lower triangle: let's see the coordinates. The lower triangle has a point at \((-4,-2)\)? No, the lower triangle's top point (corresponding to \(J\)) is at \((-2,-4)\)? Wait, maybe I mixed up the rotation. Wait, maybe the rotation is \(90^\circ\) clockwise? No, the problem says counterclockwise. Wait, let's check \(K(-2,2)\). Using the rule \((x,y)\to(-y,x)\): \(x=-2\), \(y = 2\). So \(-y=-2\), \(x=-2\). So \(K'(-2,-2)\)? But in the lower triangle, \(K\) is at \((-2,-2)\)? Wait, the lower triangle has a point at \((-2,-2)\) and \((-4,-2)\) and \((-2,-4)\). Wait, let's check \(L(-4,2)\). Using the rule: \(x=-4\), \(y = 2\). So \(-y=-2\), \(x=-4\). So \(L'(-2,-4)\)? Wait, no, \(x=-4\), so \(y'=x=-4\)? Wait, no, the rotation matrix gives \(\begin{pmatrix}-y\\x\end{pmatrix}\), so for \(L(-4,2)\), \(-y=-2\), \(x=-4\), so \(L'(-2,-4)\)? Wait, but in the lower triangle, the point corresponding to \(L\) is at \((-2,-4)\)? Wait, the lower triangle has vertices at \((-4,-2)\), \((-2,-2)\), \((-2,-4)\)? No, looking at the graph, the lower triangle: left point is \((-4,-2)\), middle point is \((-2,-2)\), bottom point is \((-2,-4)\). Wait, so let's re-express the original points:

Original \(J(-2,4)\), \(K(-2,2)\), \(L(-4,2)\).

After \(90^\circ\) counterclockwise rotation:

For \(J(-2,4)\): \((x,y)\to(-y,x)=(-4,-2)\)? But the lower triangle's left point is \((-4,-2)\), middle is \((-2,-2)\), bottom is \((-2,-4)\). Wait, maybe I got the rotation direction wrong. Wait, maybe it's \(90^\circ\) clockwise? The rule for \(90^\circ\) clockwise is \((x,y)\to(y,-x)\). Let's try that. For \(J(-2,4)\): \((4,2)\)? No, that doesn't match. Wait, no, the lower triangle is a rotation of the upper triangle. Let's see the upper triangle: \(J(-2,4)\), \(K(-2,2)\), \(L(-4,2)\). So it's a right triangle with vertical leg from \(K\) to \(J\) (length 2) and horizontal leg from \(L\) to \(K\) (length 2). The lower triangle: let's see the coordinates. The lower triangle has a vertical leg from \((-2,-2)\) to \((-2,-4)\) (length 2) and horizontal leg from \((-4,-2)\) to \((-2,-2)\) (length 2). So the rotation: if we rotate the upper triangle \(90^\circ\) counterclockwise, the vertical leg (along \(x=-2\), from \(y=2\) to \(y=4\)) would become horizontal? Wait, no. Wait, the upper triangle: \(J(-2,4)\), \(K(-2,2)\), \(L(-4,2)\). So vector \(KJ\) is \((0,2)\), vector \(KL\) is \((-2,0)\). After \(90^\circ\) counterclockwise rotation, vector \(KJ\) becomes \((-2,0)\) (since rotating \((0,2)\) 90 CCW is \((-2,0)\)) and vector \(KL\) becomes \((0,-2)\) (rotating \((-2,0)\) 90 CCW is \((0,-2)\)). So the new triangle would have \(K'\) at some point, \(J'\) at \(K' + (-2,0)\), and \(L'\) at \(K' + (0,-2)\). Let's find \(K'\). Original \(K(-2,2)\). Using the rotation rule \((x,y)\to(-y,x)\): \((-2,2)\to(-2,-2)\). So \(K'(-2,-2)\). Then \(J'\) is \(K' + (-2,0)\)? Wait, no, original \(J\) is \(K + (0,2)\). After rotation, \(J'\) should be \(K' + (-2,0)\) (since rotating \((0,2)\) 90 CCW is \((-2,0)\)). So \(J'(-2 - 2, -2 + 0)=(-4,-2)\). And \(L\) is \(K + (-2,0)\), so after rotation, \(L'\) is \(K' + (0,-2)\) (since rotating \((-2,0)\) 90 CCW is \((0,-2)\)). So \(L'(-2 + 0, -2 - 2)=(-2,-4)\). Let's check:

- \(J(-2,4)\) rotated 90 CCW: \((-4,-2)\)
- \(K(-2,2)\) rotated 90 CCW: \((-2,-2)\)
- \(L(-4,2)\) rotated 90 CCW: \((-2,-4)\)

Yes, that matches the lower triangle's coordinates: \((-4,-2)\), \((-2,-2)\), \((-2,-4)\).

So:

- For \(J(-2,4)\): apply \((x,y)\to(-y,x)\), so \(x=-2\), \(y=4\), new \(x=-4\), new \(y=-2\) → \(J'(-4,-2)\)
- For \(K(-2,2)\): \(x=-2\), \(y=2\), new \(x=-2\), new \(y=-2\) → \(K'(-2,-2)\)
- For \(L(-4,2)\): \(x=-4\), \(y=2\), new \(x=-2\), new \(y=-4\) → \(L'(-2,-4)\)

Answer:

• \(J'\): \((-4, -2)\)
• \(K'\): \((-2, -2)\)
• \(L'\): \((-2, -4)\)