The given problem involves a triangle $ \Delta MNP $ and asks for its reflection and an explanation regarding its area. The triangle has vertices at $ M(-10, -2) $, $ N(-6, -9) $, and $ P(1, -5) $. Additionally, a line $ y = \frac{2}{3}x $ is given.**Problem Breakdown:****(a)** Draw the image of $ \Delta MNP $ after a reflection in $ y = \frac{2}{3} x $. Give the coordinates of the transformed vertices below.1. **Original Triangle $ \Delta MNP $**: - Vertex $ M $ at $ (-10, -2) $ - Vertex $ N $ at $ (-6, -9) $ - Vertex $ P $ at $ (1, -5) $2. **Reflection Line**: $ y = \frac{2}{3}x $3. **Reflection Procedure**: - To find the reflected coordinates of a point across the line $ y = \frac{2}{3}x $, one needs to use a formula or perform a geometric transformation on each point such that the points' positions are swapped symmetrically across the line.**Coordinates of the Reflected Vertices:****Using Hypothetical Reflection Technique (not calculated here):** - Reflected Vertex $ M' $ - Reflected Vertex $ N' $ - Reflected Vertex $ P' $(Exact calculations for these points require geometric manipulation techniques which can be derived using tools like matrix transformations or software that can visualize reflections).**(b)** Explain why $ \Delta M'N'P' $ must have the same area as $ \Delta MNP $.- Reflection is a type of isometry. An isometry is any transformation that preserves distances and angles.- Since area is a measurement dependent on distances (base and height), an isometric transformation such as reflection will keep the area of the original figure unchanged.- Therefore, $ \Delta M'N'P' $, being the reflected image of $ \Delta MNP $, will maintain the same area as $ \Delta MNP $.**Graph Explanation:**- The graph is a standard Cartesian coordinate system with the x-axis and y-axis labeled.- The line \( y =

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6. On the grid below AMNP is plotted with vertices at M(-10,– 2). N(-6,-9) and P(1, -5). The line y=-x is also drawn. (a) Draw the image of AMNP after a reflection in y=-r. Give the coordinates of the transformed vertices below. (b) Explain why AM'N'P" must have the same area as AMNP.

6. On the grid below AMNP is plotted with vertices at M(-10,– 2). N(-6,-9) and P(1, -5). The line y=-x is also drawn. (a) Draw the image of AMNP after a reflection in y=-r. Give the coordinates of the transformed vertices below. (b) Explain why AM'N'P" must have the same area as AMNP.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter9: Real Numbers And Right Triangles

Section9.5: The Distance And Midpoint Formulas

Problem 2C

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Question

The given problem involves a triangle $ \Delta MNP $ and asks for its reflection and an explanation regarding its area. The triangle has vertices at $ M(-10, -2) $, $ N(-6, -9) $, and $ P(1, -5) $. Additionally, a line $ y = \frac{2}{3}x $ is given.

**Problem Breakdown:**

**(a)** Draw the image of $ \Delta MNP $ after a reflection in $ y = \frac{2}{3} x $. Give the coordinates of the transformed vertices below.

1. **Original Triangle $ \Delta MNP $**:
- Vertex $ M $ at $ (-10, -2) $
- Vertex $ N $ at $ (-6, -9) $
- Vertex $ P $ at $ (1, -5) $

2. **Reflection Line**: $ y = \frac{2}{3}x $

3. **Reflection Procedure**:
- To find the reflected coordinates of a point across the line $ y = \frac{2}{3}x $, one needs to use a formula or perform a geometric transformation on each point such that the points' positions are swapped symmetrically across the line.

**Coordinates of the Reflected Vertices:**

**Using Hypothetical Reflection Technique (not calculated here):**
- Reflected Vertex $ M' $
- Reflected Vertex $ N' $
- Reflected Vertex $ P' $

(Exact calculations for these points require geometric manipulation techniques which can be derived using tools like matrix transformations or software that can visualize reflections).

**(b)** Explain why $ \Delta M'N'P' $ must have the same area as $ \Delta MNP $.

- Reflection is a type of isometry. An isometry is any transformation that preserves distances and angles.
- Since area is a measurement dependent on distances (base and height), an isometric transformation such as reflection will keep the area of the original figure unchanged.
- Therefore, $ \Delta M'N'P' $, being the reflected image of $ \Delta MNP $, will maintain the same area as $ \Delta MNP $.

**Graph Explanation:**
- The graph is a standard Cartesian coordinate system with the x-axis and y-axis labeled.
- The line \( y =

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