Question

plot points L(-9,-6) M(-2,-6) N(4,-2) O(-3,-2)

which segments are parallel?
which segments are perpendicular?
what is the name of the figure?
find the perimeter?
find the area/

Answer 1

To solve this problem, let's analyze the points \( L(-9,-6) \), \( M(-2,-6) \), \( N(4,-2) \), and \( O(-3,-2) \).

1. **Plotting the Points**:
- \( L(-9,-6) \)
- \( M(-2,-6) \)
- \( N(4,-2) \)
- \( O(-3,-2) \)

These points can be plotted on the Cartesian plane.

2. **Finding the Slope of Segments**:
We will calculate the slopes of the segments to determine which are parallel and which are perpendicular.

- **Segment \( LM \)**:
\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - (-6)}{-2 - (-9)} = \frac{0}{7} = 0
\]
- **Segment \( MN \)**:
\[
\text{slope} = \frac{-2 - (-6)}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}
\]
- **Segment \( NO \)**:
\[
\text{slope} = \frac{-2 - (-2)}{-3 - 4} = \frac{0}{-7} = 0
\]
- **Segment \( OL \)**:
\[
\text{slope} = \frac{-6 - (-2)}{-9 - (-3)} = \frac{-4}{-6} = \frac{2}{3}
\]

3. **Determining parallel and perpendicular segments**:
- **Parallel Segments**: \( LM \) and \( NO \) (both have slope 0; they are horizontal lines).
- **Parallel Segments**: \( MN \) and \( OL \) (both have the slope of \( \frac{2}{3} \)).
- **Perpendicular Segments**: No segments are perpendicular since there are no segments that have slopes that multiply to \(-1\) (the slopes of \( LM \) and \( NO \) = 0 don't meet this criteria).

4. **Naming the Figure**:
The figure formed by the points \( L, M, N, O \) is a trapezoid since it has at least one pair of parallel sides (horizontal segments \( LM \) and \( NO \)) and the other two sides are non-parallel.

5. **Finding the Perimeter**:
We can calculate the lengths of the segments using the distance formula:
- **Length \( LM \)**:
\[
LM = \sqrt{(-2 - (-9))^2 + (-6 - (-6))^2} = \sqrt{7^2} = 7
\]
- **Length \( MN \)**:
\[
MN = \sqrt{(4 - (-2))^2 + (-2 - (-6))^2} = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.211
\]
- **Length \( NO \)**:
\[
NO = \sqrt{(-3 - 4)^2 + (-2 - (-2))^2} = \sqrt{(-7)^2} = 7
\]
- **Length \( OL \)**:
\[
OL = \sqrt{(-9 - (-3))^2 + (-6 - (-2))^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.211
\]

Now, sum the lengths:
\[
\text{Perimeter} = LM + MN + NO + OL = 7 + 2\sqrt{13} + 7 + 2\sqrt{13} = 14 + 4\sqrt{13} \approx 14 + 4(3.605) \approx 14 + 14.42 = 28.42
\]

6. **Finding the Area**:
The area \( A \) of a trapezoid can be calculated using the formula:
\[
A = \frac{1}{2} (b_1 + b_2) h
\]
Where \( b_1 \) and \( b_2 \) are the lengths of the two bases and \( h \) is the height.
- Bases \( b_1 = LM = 7 \), \( b_2 = NO = 7 \)
- Height can be determined from the \( y \)-coordinates of the lines. The vertical distance from \( y = -6 \) to \( y = -2 \) is \( 4 \).

Therefore, the area \( A \):
\[
A = \frac{1}{2} (7 + 7) (4) = \frac{1}{2} (14)(4) = 28
\]

### Summary:
- **Parallel Segments**: \( LM \parallel NO \) and \( MN \parallel OL \)
- **Perpendicular Segments**: None
- **Name of Figure**: Trapezoid
- **Perimeter**: \( 14 + 4\sqrt{13} \) or approx. 28.42
- **Area**: \( 28 \) square units