https://file.aiquickdraw.com/mathbot/file/1726589302371uf5xvoue.jpg

Answer all them list Them in order and show work in a simple understandable way

I have displayed the image for analysis. Now, let's address each of the quiz questions in order.

### 1. **Give two names for the plane.**
- Based on the diagram, the plane can be named using three non-collinear points in the plane.
  - Two possible names are:
    - Plane **ABC**
    - Plane **ABD**

### 2. **Name three collinear points.**
- Collinear points are points that lie on the same straight line.
  - Points **A**, **B**, and **C** are collinear.

### 3. **Name three coplanar points.**
- Coplanar points are points that lie on the same plane.
  - Points **A**, **B**, and **D** are coplanar.

### 4. **Name three points.**
- You can select any three points from the diagram. For example:
  - Points **A**, **B**, and **D**.

### 5. **Name one ray.**
- A ray starts from one point and goes infinitely in one direction.
  - Ray **AB** (starting at A and passing through B).

### 6. **Name two lines.**
- From the diagram, two lines can be named:
  - Line **AC**
  - Line **BD**

### 7. **Name one line segment.**
- A line segment has two endpoints.
  - Line segment **AB**.

### 8. **Sketch the figure described: AB and AC.**
- This requires drawing the two rays **AB** and **AC**, which share the same starting point A and move in different directions.

### 9. **Plane C and plane D intersecting at XY.**
- This suggests that two planes intersect at line **XY**. You can visualize this as a line where two flat surfaces meet.

### 10. **Find BD.**
- Based on the work shown on the page, BD is given as 17.

### 11. **The endpoints of \(\overline{QR}\) are Q(1,6) and R(7,3). Find the coordinates of the midpoint M.**
- To find the midpoint \(M(x,y)\) of a segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\), use the midpoint formula:
  \[
  M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
  \]
  Substituting:
  \[
  M = \left(\frac{1 + 7}{2}, \frac{6 + 3}{2}\right) = (4, 4.5)
  \]
  So, the midpoint \(M\) is \((4, 4.5)\).

### 12. **Find the distance between the two points S(-5, -2) and T(7, 4).**
- Use the distance formula:
  \[
  d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  \]
  Substituting:
  \[
  d = \sqrt{(7 - (-5))^2 + (4 - (-2))^2} = \sqrt{(12)^2 + (6)^2} = \sqrt{144 + 36} = \sqrt{180}
  \]
  So, \(d = \sqrt{180} = 13.42\) (rounded to two decimal places).

### 13. **Identify the segment bisector of \(\overline{QR}\).**
- The bisector of a segment divides it into two equal parts. Based on the diagram, the bisector seems to be a line passing through the midpoint of \(\overline{QR}\).

### 14. **Find the coordinates of endpoint H if the midpoint of \(\overline{GH}\) is M(4, -3) and G is (2, 4).**
- Use the midpoint formula to find the missing endpoint \(H(x_2, y_2)\). We know the midpoint formula is:
  \[
  M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
  \]
  For the x-coordinate:
  \[
  4 = \frac{2 + x_2}{2} \implies 8 = 2 + x_2 \implies x_2 = 6
  \]
  For the y-coordinate:
  \[
  -3 = \frac{4 + y_2}{2} \implies -6 = 4 + y_2 \implies y_2 = -10
  \]
  Therefore, the coordinates of \(H\) are \((6, -10)\).

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Feel free to ask for more details or clarification!

Here are 5 related questions to deepen your understanding:
1. What is the difference between a ray and a line?
2. How do you determine if points are collinear?
3. What is the difference between coplanar and collinear points?
4. How do you find the equation of a line given two points?
5. What is the relationship between the segment bisector and the midpoint?

**Tip:** Always double-check your substitution in formulas like the distance or midpoint formulas to avoid simple mistakes.

Answer all them list them in order and show work in a simple understandable way

Solve geometry problems involving line segments, midpoints, and the distance formula. Understand core concepts like collinearity, ray and line naming, and segment bisectors with a step-by-step solution approach. Suitable for middle school students (Grades 7-9).

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