lines ( l ) and ( m ) are perpendicular ( m perp l ). a point ( q ) has this property: rotating ( q ) 180 degrees using center ( p ) has the same effect as reflecting ( q ) over line ( m ).
a. give two possible locations of ( q ).
b. do all points in the plane have this property? explain how you know.

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Analyze Point on Line $ m $
A point on line $ m $ (excluding $ P $): Let's take a point $ Q $ on line $ m $, say $ Q $ is above $ P $ on $ m $. Rotating $ Q $ 180° around $ P $ will map it to a point below $ P $ on $ m $. Reflecting $ Q $ over $ m $ (since $ Q $ is on $ m $) leaves it unchanged? Wait, no—wait, $ m $ is the line of reflection. Wait, no, the rotation is 180° around $ P $, and reflection over $ m $. Let's think about the composition. A 180° rotation around $ P $ is equivalent to two reflections over lines through $ P $ (perpendicular). Since $ l \perp m $, let's denote $ l $ and $ m $ as perpendicular lines through $ P $. A 180° rotation around $ P $ is the same as reflecting over $ l $ then over $ m $ (or vice versa). But the problem says rotating 180° around $ P $ is the same as reflecting over $ m $. Wait, that would mean that reflecting over $ l $ then over $ m $ is the same as reflecting over $ m $, which would imply reflecting over $ l $ is the identity, which is not true. Wait, maybe I misread. Wait, the problem says "rotating $ Q $ 180 degrees using center $ P $ has the same effect as reflecting $ Q $ over line $ m $". So $ R_{P,180^\circ}(Q) = r_m(Q) $, where $ R_{P,180^\circ} $ is 180° rotation around $ P $, and $ r_m $ is reflection over $ m $. Let's recall that $ R_{P,180^\circ}(Q) $ is the point $ Q' $ such that $ P $ is the midpoint of $ QQ' $. $ r_m(Q) $ is the point $ Q'' $ such that $ m $ is the perpendicular bisector of $ QQ'' $. So we need $ P $ to be the midpoint of $ QQ'' $ and $ m $ to be the perpendicular bisector of $ QQ'' $. Therefore, $ QQ'' $ must be perpendicular to $ m $, and $ P $ is the midpoint. But if $ QQ'' $ is perpendicular to $ m $, then $ QQ'' $ is parallel to $ l $ (since $ l \perp m $). Wait, maybe points on line $ l $? Let's take a point $ Q $ on line $ l $, say to the right of $ P $ on $ l $. Rotating $ Q $ 180° around $ P $ maps it to the left of $ P $ on $ l $. Reflecting $ Q $ over $ m $: since $ l \perp m $, the reflection of a point on $ l $ over $ m $ will be a point such that $ m $ is the perpendicular bisector. Wait, maybe $ P $ itself? If $ Q = P $, rotating 180° around $ P $ leaves it at $ P $, reflecting over $ m $ also leaves it at $ P $. So $ Q = P $ is a solution. Another point: let's take a point $ Q $ such that $ P $ is the midpoint of $ Q $ and $ r_m(Q) $, and also $ R_{P,180^\circ}(Q) = r_m(Q) $. So $ R_{P,180^\circ}(Q) = r_m(Q) \implies $ midpoint of $ Q $ and $ r_m(Q) $ is $ P $, and $ m $ is the perpendicular bisector of $ Qr_m(Q) $. Therefore, $ Qr_m(Q) $ is perpendicular to $ m $, so $ Qr_m(Q) $ is parallel to $ l $ (since $ l \perp m $). So $ Q $ lies on a line parallel to $ l $, and $ P $ is the midpoint of $ Q $ and $ r_m(Q) $. But also, $ R_{P,180^\circ}(Q) = r_m(Q) \implies $ $ r_m(Q) = - \vec{PQ} + P $ (vectorially, 180° rotation). And $ r_m(Q) $ is the reflection over $ m $, so if $ m $ is the y-axis (let's set coordinate system: $ P $ at (0,0), $ m $ as y-axis, $ l $ as x-axis). Then reflection over $ m $ (y-axis) of $ (x,y) $ is $ (-x,y) $. 180° rotation around $ P(0,0) $ of $ (x,y) $ is $ (-x,-y) $. The problem says these are equal: $ (-x,-y) = (-x,y) \implies -y = y \implies y = 0 $. Wait, that's a key insight! So in coordinates, let $ P = (0,0) $, $ m $ (y-axis), $ l $ (x-axis). Then 180° rotation of $ (x,y) $ is $ (-x,-y) $, reflection over $ m $ (y-axis) is $ (-x,y) $. Setting them equal: $ (-x,-y) = (-x,y) \implies y = 0 $. So $ Q $ must lie on the x-axis (line $ l $)! Wait, that makes sense. So if $ Q $ is on line $ l $, then 180° rotation around $ P $ (which is on $ l $): for $ Q = (a,0) $, 180° rotation is $ (-a,0) $. Reflection over $ m $ (y-axis) of $ (a,0) $ is $ (-a,0) $. So they are equal! Ah, so points on line $ l $ satisfy this. Also, point $ P $ itself: $ Q = P $, rotation 180° leaves it at $ P $, reflection over $ m $ also leaves it at $ P $. So two possible locations: $ P $ itself, and any point on line $ l $ (e.g., $ Q $ on $ l $ to the right of $ P $, or on $ l $ to the left, or $ P $).

So two possible locations: $ Q = P $, and $ Q $ on line $ l $ (e.g., $ Q $ is a point on line $ l $, such as the point to the right of $ P $ on $ l $, or the point to the left, or $ P $).

## Step2: Confirm with Coordinates
Let’s set $ P = (0,0) $, $ m $ as the y-axis ($ x=0 $), $ l $ as the x-axis ($ y=0 $). Let $ Q = (a,0) $ on $ l $. Rotating $ Q $ 180° around $ P $: $ (a,0) \to (-a,0) $. Reflecting $ Q $ over $ m $ (y-axis): $ (a,0) \to (-a,0) $. So they are equal. Now take $ Q = (0,b) $ on $ m $. Rotating 180° around $ P $: $ (0,b) \to (0,-b) $. Reflecting over $ m $ (y-axis): $ (0,b) \to (0,b) $ (since it's on the y-axis). These are not equal unless $ b = 0 $ (i.e., $ Q = P $). So only points on $ l $ (and $ P $) satisfy? Wait, but $ P $ is on $ l $. So two possible locations: $ P $ itself, and a point on line $ l $ (e.g., $ (1,0) $ if $ P $ is $ (0,0) $, or $ (-1,0) $, or $ P $).

# Answer (a):
Two possible locations of $ Q $ are:
- $ Q = P $ (the intersection point of $ l $ and $ m $).
- Any point on line $ l $ (e.g., a point to the right of $ P $ on $ l $, or to the left of $ P $ on $ l $).

### Part (b)
# Explanation:
## Step1: Analyze a Point Not on $ l $
Take a point $ Q $ not on line $ l $ (and not $ P $). Let’s use coordinates: $ P = (0,0) $, $ m $ (y-axis), $ l $ (x-axis). Let $ Q = (a,b) $ where $ b
eq 0 $ (so not on $ l $). Rotating $ Q $ 180° around $ P $: $ (a,b) \to (-a,-b) $. Reflecting $ Q $ over $ m $ (y-axis): $ (a,b) \to (-a,b) $. For these to be equal, we need $ -a,-b = -a,b \implies -b = b \implies b = 0 $. But we chose $ b
eq 0 $, so they are not equal. Thus, points not on line $ l $ (and not $ P $) do not satisfy the property.

## Step2: Conclusion
Only points on line $ l $ (including $ P $) have the property that rotating 180° around $ P $ is the same as reflecting over $ m $. Therefore, not all points in the plane have this property.

# Answer (b):
No, not all points in the plane have this property. For a point $ Q $ not on line $ l $ (e.g., a point with a non - zero y - coordinate in the coordinate system where $ P=(0,0) $, $ l $ is the x - axis, and $ m $ is the y - axis), a 180° rotation around $ P $ and a reflection over $ m $ produce different results (the y - coordinates of the images differ). Only points on line $ l $ (including $ P $) satisfy the property.

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Analyze Point on Line \( m \)

Step1: Analyze a Point Not on \( l \)

Step2: Conclusion

Answer:

Part (b)