name: date: 1. in δtuv, y is the centroid. a) if yw = 9, find ty and tw. b) if vx = 9, find vy and yx. 2. for δabc, is the red segment a median, an altitude, or neither? explain. a) b) 3. name the centroid of δprq. 4. name the orthocenter of δxyz. 5. find the coordinates of the orthocenter of δabc. a(0, 0) b(4, 0) c(4, 2)

Question

Accepted Answer

### Problem 1a)
# Explanation:
## Step1: Recall centroid property (median ratio)
The centroid $ Y $ divides each median into a ratio of $ 2:1 $, with the longer segment being closer to the vertex. For median $ TW $ (where $ Y $ is centroid), $ TY:YW = 2:1 $. Given $ YW = 9 $.
## Step2: Calculate $ TY $
Since $ TY = 2 \times YW $, substitute $ YW = 9 $: $ TY = 2 \times 9 = 18 $.
## Step3: Calculate $ TW $
$ TW = TY + YW = 18 + 9 = 27 $.

# Answer:
$ TY = 18 $, $ TW = 27 $

### Problem 1b)
# Explanation:
## Step1: Recall centroid property (median ratio)
For median $ VX $ (centroid $ Y $), $ VY:YX = 2:1 $, so total parts $ 2 + 1 = 3 $. Given $ VX = 9 $.
## Step2: Calculate $ VY $
$ VY = \frac{2}{3} \times VX $, substitute $ VX = 9 $: $ VY = \frac{2}{3} \times 9 = 6 $.
## Step3: Calculate $ YX $
$ YX = \frac{1}{3} \times VX = \frac{1}{3} \times 9 = 3 $.

# Answer:
$ VY = 6 $, $ YX = 3 $

### Problem 2a) (Assuming red segment is from $ A $ to $ BC $ with right angle at $ A $ to $ BC $)
# Explanation:
- **Median**: Connects vertex to midpoint of opposite side.
- **Altitude**: Perpendicular segment from vertex to opposite side (or its extension).
The red segment is perpendicular to $ BC $ (right angle symbol), so it is an altitude (since it’s perpendicular from $ A $ to $ BC $).

# Answer:
Altitude; it is perpendicular to $ BC $ (forms a right angle), so it’s an altitude.

### Problem 2b) (Red segment is from $ B $ to $ AC $ with right angle at $ AC $)
# Explanation:
- **Altitude**: Perpendicular to $ AC $ (right angle symbol), so it is an altitude (perpendicular from $ B $ to $ AC $). Also, since $ A $ and $ C $ are endpoints, if $ B $ to $ AC $ is perpendicular and $ AC $ is a side, it’s an altitude. If it also hits the midpoint, but here the right angle confirms perpendicularity (altitude definition).

# Answer:
Altitude; it is perpendicular to $ AC $ (forms a right angle), so it’s an altitude.

### Problem 3
# Explanation:
The centroid of a triangle is the intersection of its medians. In $ \triangle PRQ $, the medians are $ PL $ (midpoint of $ RQ $), $ RM $ (midpoint of $ PQ $), and $ QN $ (midpoint of $ PR $). Their intersection is $ M $ (assuming $ M $ is the centroid in the diagram).

# Answer:
$ M $ (the intersection of the three medians of $ \triangle PRQ $)

### Problem 4
# Explanation:
The orthocenter of a triangle is the intersection of its altitudes. In $ \triangle XYZ $, the altitudes are perpendicular segments from each vertex to the opposite side. The diagram shows $ V $ as the intersection of altitudes (e.g., $ YW $ is an altitude, $ XU $ is an altitude, intersecting at $ V $).

# Answer:
$ V $ (the intersection of the altitudes of $ \triangle XYZ $)

### Problem 5
# Explanation:
## Step1: Identify sides and slopes (find altitudes)
- $ A(0,0) $, $ B(4,0) $, $ C(4,2) $.
- Side $ AB $: horizontal (slope $ 0 $), so altitude from $ C $ to $ AB $ is vertical (perpendicular to horizontal). Since $ AB $ is on $ y = 0 $, the altitude from $ C(4,2) $ to $ AB $ is vertical line $ x = 4 $ (same $ x $-coordinate as $ C $, perpendicular to $ AB $).
- Side $ BC $: vertical ( $ x = 4 $ ), so altitude from $ A $ to $ BC $ is horizontal (perpendicular to vertical). Since $ BC $ is $ x = 4 $, the altitude from $ A(0,0) $ to $ BC $ is horizontal line $ y = 0 $ (same $ y $-coordinate as $ A $, perpendicular to $ BC $).
## Step2: Find intersection of altitudes
Altitude from $ C $: $ x = 4 $; Altitude from $ A $: $ y = 0 $. Intersection is $ (4, 0) $.

# Answer:
$ (4, 0) $

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QUESTION IMAGE

Question

Explanation:

Problem 1a)

Step1: Recall centroid property (median ratio)

Step2: Calculate \( TY \)

Step3: Calculate \( TW \)

Step1: Recall centroid property (median ratio)

Step2: Calculate \( VY \)

Step3: Calculate \( YX \)

Answer:

Problem 1b)