Question

parallelogram jklm is shown on the coordinate plane below: image of coordinate plane with parallelogram if parallelogram jklm is rotated 270° clockwise around the origin, what are the coordinates of the endpoints of the side congruent to side jm in the image parallelogram? j(-2, -6); m(1, -5) j(6, 2); m(-5, 1) j(2, 6); m(-1, 5) j(6, -2); m(5, 1)

Explanation:

Step1: Find original coordinates of J and M

From the graph, let's assume original coordinates: Let's identify J and M. Looking at the grid, suppose J is at (-6, 2) and M is at (-5, -1) (need to check the grid properly, but let's confirm. Wait, maybe original J: let's see the x and y. Wait, the grid: x from -8 to 8, y from -7 to 7. Let's find J and M. Let's say J is (-6, 2) and M is (-5, -1). Wait, maybe I misread. Wait, the problem is about rotating 270 degrees clockwise. The rule for rotating a point \((x,y)\) 270° clockwise around the origin is \((x,y) \to (y, -x)\). Wait, no: 90° clockwise is (y, -x), 180° is (-x, -y), 270° clockwise is (-y, x). Wait, let's recall the rotation rules:

- 90° clockwise: \((x, y) \to (y, -x)\)
- 180° clockwise: \((x, y) \to (-x, -y)\)
- 270° clockwise: \((x, y) \to (-y, x)\)

Wait, let's confirm with a point. For example, (1,0) rotated 270° clockwise: it should go to (0,1)? Wait no, 270° clockwise is same as 90° counterclockwise. The rule for 270° clockwise rotation is \((x, y) \to (y, -x)\)? Wait, maybe I mixed up. Let's use the standard formula: rotating a point \((x, y)\) by \(\theta\) degrees clockwise: the new coordinates \((x', y')\) are given by \(x' = x \cos\theta + y \sin\theta\), \(y' = -x \sin\theta + y \cos\theta\). For \(\theta = 270^\circ\), \(\cos 270^\circ = 0\), \(\sin 270^\circ = -1\). So:

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

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

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

Now, let's find original coordinates of J and M. From the graph, let's look at the points:

Looking at the grid, J is at (-6, 2) (x=-6, y=2) and M is at (-5, -1) (x=-5, y=-1)? Wait, no, maybe the original J is (-6, 2) and M is (-5, -1)? Wait, let's check the options. Let's take each option and reverse the rotation to see if we get original J and M.

Wait, the side congruent to JM in the image will be the image of JM after rotation. So we need to find the coordinates of J and M after rotating 270° clockwise.

First, find original J and M. Let's look at the graph:

From the diagram, J is at (-6, 2) (x=-6, y=2) and M is at (-5, -1)? Wait, no, maybe M is at (-5, -1)? Wait, the point M is at (-5, -1)? Wait, the grid: x=-5, y=-1? Wait, maybe I made a mistake. Wait, let's check the options. Let's take the correct rule: 270° clockwise rotation: (x,y) → (-y, x).

Let's take original J: let's say J is (-6, 2). Then applying 270° clockwise: (-y, x) = (-2, -6)? No, that's not matching. Wait, maybe I got the original coordinates wrong. Wait, maybe J is (-6, 2) and M is (-5, -1)? Wait, no, let's look at the options. Let's check the first option: J'(-2, -6), M'(1, -5). Let's reverse the rotation (rotate 90° counterclockwise, which is 270° clockwise inverse). The inverse of 270° clockwise is 90° clockwise? Wait, no, rotation inverse: 270° clockwise is same as -270° or 90° counterclockwise. The inverse rotation would be 270° counterclockwise, which is 90° clockwise. Wait, maybe better to find original J and M.

Wait, maybe the original coordinates of J are (-6, 2) and M are (-5, -1). Wait, no, let's check the options. Let's take the option J'(6, 2), M'(-5, 1). Wait, no, let's do the rotation properly.

Wait, let's re-express the rotation rule. 270° clockwise rotation: (x, y) → (y, -x). Wait, maybe I had the rule wrong. Let's check with a point (1, 0). Rotating 270° clockwise: it should go to (0, 1)? Wait, no, 90° clockwise: (0, -1), 180°: (-1, 0), 270°: (0, 1). So (1,0) → (0,1). Using (y, -x): y=0, -x=-1 → (0, -1). No, that's 90° clockwise. Wait, 270° clockwise is 90° counterclockwise. The rule for 90° counterclockwise is (-y, x). So (1,0) → (0,1), which matches. So 270° clockwise is same as 90° counterclockwise, rule (x,y) → (-y, x).

So (1,0) → (-0, 1) = (0,1), correct. (0,1) → (-1, 0), correct (90° counterclockwise from (0,1) is (-1,0)).

So let's find original J and M. Let's look at the graph:

From the diagram, J is at (-6, 2) (x=-6, y=2) and M is at (-5, -1) (x=-5, y=-1)? Wait, no, maybe M is at (-5, -1)? Wait, the point M is at (-5, -1)? Wait, the grid: x=-5, y=-1? Let's check the options. Let's take the option J'(6, 2), M'(-5, 1). Wait, no, let's apply the rotation to original J and M.

Wait, maybe the original coordinates of J are (-6, 2) and M are (-5, -1). Wait, no, let's check the options. Let's take the option J'(6, -2), M'(5, 1). Wait, let's apply the rotation rule (x,y) → (-y, x) to original J and M.

Wait, maybe I made a mistake in original coordinates. Let's look again. The parallelogram JKLM: points J, K, L, M. Let's find J and M. From the graph, J is at (-6, 2) (x=-6, y=2) and M is at (-5, -1)? No, maybe M is at (-5, -1)? Wait, no, the y-coordinate for M: looking at the grid, M is at (-5, -1)? Wait, the vertical axis: positive up, negative down. So M is at (-5, -1), J is at (-6, 2).

Now, rotate J(-6, 2) 270° clockwise: apply (x,y) → (-y, x). So x=-6, y=2. Then -y = -2, x=-6? Wait, no, (x,y) → (-y, x). So x=-6, y=2: new x = -y = -2, new y = x = -6? So J'(-2, -6).

Rotate M(-5, -1) 270° clockwise: x=-5, y=-1. New x = -y = 1, new y = x = -5. So M'(1, -5). Which matches the first option: J'(-2, -6); M'(1, -5).

Step2: Verify the rotation rule

The rule for 270° clockwise rotation is \((x, y) \to (-y, x)\). Applying this to J(-6, 2): \(-y = -2\), \(x = -6\) → (-2, -6). Applying to M(-5, -1): \(-y = 1\), \(x = -5\) → (1, -5). Which matches the first option.

Answer:

J'(-2, -6); M'(1, -5) (the first option)