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Sagot :
To determine the type of quadrilateral and its slopes, let's proceed with step-by-step calculations.
### Step 1: Determine the slopes of each side
To find the slope of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], we use the formula:
[tex]\[ \text{slope} = \frac{(y_2 - y_1)}{(x_2 - x_1)} \][/tex]
#### Slope of segment AB:
Points [tex]\(A(5, 6)\)[/tex] and [tex]\(B(13, 6)\)[/tex]
[tex]\[ \text{slope}_{AB} = \frac{(6 - 6)}{(13 - 5)} = \frac{0}{8} = 0 \][/tex]
#### Slope of segment BC:
Points [tex]\(B(13, 6)\)[/tex] and [tex]\(C(11, 2)\)[/tex]
[tex]\[ \text{slope}_{BC} = \frac{(2 - 6)}{(11 - 13)} = \frac{-4}{-2} = 2 \][/tex]
#### Slope of segment CD:
Points [tex]\(C(11, 2)\)[/tex] and [tex]\(D(1, 2)\)[/tex]
[tex]\[ \text{slope}_{CD} = \frac{(2 - 2)}{(1 - 11)} = \frac{0}{-10} = 0 \][/tex]
#### Slope of segment DA:
Points [tex]\(D(1, 2)\)[/tex] and [tex]\(A(5, 6)\)[/tex]
[tex]\[ \text{slope}_{DA} = \frac{(6 - 2)}{(5 - 1)} = \frac{4}{4} = 1 \][/tex]
So, the slopes of the sides are:
- [tex]\( \text{slope}_{AB} = 0 \)[/tex]
- [tex]\( \text{slope}_{BC} = 2 \)[/tex]
- [tex]\( \text{slope}_{CD} = 0 \)[/tex]
- [tex]\( \text{slope}_{DA} = 1 \)[/tex]
### Step 2: Analyze the slopes to determine the type of the quadrilateral
Given the slopes:
- [tex]\( \text{slope}_{AB} = 0 \)[/tex]
- [tex]\( \text{slope}_{BC} = 2 \)[/tex]
- [tex]\( \text{slope}_{CD} = -0 \)[/tex]
- [tex]\( \text{slope}_{DA} = 1 \)[/tex]
We see that:
- Opposite sides [tex]\(AB\)[/tex] and [tex]\(CD\)[/tex] are horizontal (slopes of 0 and -0 respectively).
- The other pair of sides, [tex]\(BC\)[/tex] and [tex]\(DA\)[/tex], have different slopes (2 and 1 respectively).
Since not all slopes are equal, and because the slopes do not indicate pairs of parallel and equal-length sides (which would verify if it is a rectangle or square), the quadrilateral cannot be a rectangle or a square.
### Final Conclusion:
The quadrilateral with given vertices [tex]\(A(5, 6)\)[/tex], [tex]\(B(13, 6)\)[/tex], [tex]\(C(11, 2)\)[/tex], and [tex]\(D(1, 2)\)[/tex] is an Other Quadrilateral.
### Step 1: Determine the slopes of each side
To find the slope of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], we use the formula:
[tex]\[ \text{slope} = \frac{(y_2 - y_1)}{(x_2 - x_1)} \][/tex]
#### Slope of segment AB:
Points [tex]\(A(5, 6)\)[/tex] and [tex]\(B(13, 6)\)[/tex]
[tex]\[ \text{slope}_{AB} = \frac{(6 - 6)}{(13 - 5)} = \frac{0}{8} = 0 \][/tex]
#### Slope of segment BC:
Points [tex]\(B(13, 6)\)[/tex] and [tex]\(C(11, 2)\)[/tex]
[tex]\[ \text{slope}_{BC} = \frac{(2 - 6)}{(11 - 13)} = \frac{-4}{-2} = 2 \][/tex]
#### Slope of segment CD:
Points [tex]\(C(11, 2)\)[/tex] and [tex]\(D(1, 2)\)[/tex]
[tex]\[ \text{slope}_{CD} = \frac{(2 - 2)}{(1 - 11)} = \frac{0}{-10} = 0 \][/tex]
#### Slope of segment DA:
Points [tex]\(D(1, 2)\)[/tex] and [tex]\(A(5, 6)\)[/tex]
[tex]\[ \text{slope}_{DA} = \frac{(6 - 2)}{(5 - 1)} = \frac{4}{4} = 1 \][/tex]
So, the slopes of the sides are:
- [tex]\( \text{slope}_{AB} = 0 \)[/tex]
- [tex]\( \text{slope}_{BC} = 2 \)[/tex]
- [tex]\( \text{slope}_{CD} = 0 \)[/tex]
- [tex]\( \text{slope}_{DA} = 1 \)[/tex]
### Step 2: Analyze the slopes to determine the type of the quadrilateral
Given the slopes:
- [tex]\( \text{slope}_{AB} = 0 \)[/tex]
- [tex]\( \text{slope}_{BC} = 2 \)[/tex]
- [tex]\( \text{slope}_{CD} = -0 \)[/tex]
- [tex]\( \text{slope}_{DA} = 1 \)[/tex]
We see that:
- Opposite sides [tex]\(AB\)[/tex] and [tex]\(CD\)[/tex] are horizontal (slopes of 0 and -0 respectively).
- The other pair of sides, [tex]\(BC\)[/tex] and [tex]\(DA\)[/tex], have different slopes (2 and 1 respectively).
Since not all slopes are equal, and because the slopes do not indicate pairs of parallel and equal-length sides (which would verify if it is a rectangle or square), the quadrilateral cannot be a rectangle or a square.
### Final Conclusion:
The quadrilateral with given vertices [tex]\(A(5, 6)\)[/tex], [tex]\(B(13, 6)\)[/tex], [tex]\(C(11, 2)\)[/tex], and [tex]\(D(1, 2)\)[/tex] is an Other Quadrilateral.
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