Answer :
To prove that quadrilateral PART is a rectangle using coordinate geometry, we need to show two things:
1. Opposite sides are equal in length (PARALLELOGRAM): This ensures that the figure is a parallelogram.
2. One angle is 90 degrees (RIGHT ANGLE): This ensures that it is indeed a rectangle.
### Step 1: Calculate the lengths of all sides using the distance formula
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let’s calculate the lengths of [tex]\(\overline{PA}\)[/tex], [tex]\(\overline{AR}\)[/tex], [tex]\(\overline{RT}\)[/tex], and [tex]\(\overline{TP}\)[/tex]:
1. Length of [tex]\(\overline{PA}\)[/tex]:
[tex]\[ PA = \sqrt{(1 - (-5))^2 + (-4 - 6)^2} = \sqrt{(1 + 5)^2 + (-4 - 6)^2} = \sqrt{6^2 + (-10)^2} = \sqrt{36 + 100} = \sqrt{136} \][/tex]
2. Length of [tex]\(\overline{AR}\)[/tex]:
[tex]\[ AR = \sqrt{(-5 - 0)^2 + (6 - 9)^2} = \sqrt{(-5)^2 + (6 - 9)^2} = \sqrt{25 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
3. Length of [tex]\(\overline{RT}\)[/tex]:
[tex]\[ RT = \sqrt{(0 - 6)^2 + (9 - (-1))^2} = \sqrt{(0 - 6)^2 + (9 + 1)^2} = \sqrt{(-6)^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \][/tex]
4. Length of [tex]\(\overline{TP}\)[/tex]:
[tex]\[ TP = \sqrt{(6 - 1)^2 + (-1 - (-4))^2} = \sqrt{(6 - 1)^2 + (-1 + 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
### Step 2: Check if opposite sides are equal
From our calculations, we can see that:
- Length of [tex]\(\overline{PA} = \sqrt{136}\)[/tex]
- Length of [tex]\(\overline{RT} = \sqrt{136}\)[/tex]
And:
- Length of [tex]\(\overline{AR} = \sqrt{34}\)[/tex]
- Length of [tex]\(\overline{TP} = \sqrt{34}\)[/tex]
Thus, opposite sides are equal:
[tex]\[ PA = RT \][/tex]
[tex]\[ AR = TP \][/tex]
### Step 3: Calculate the slopes of the sides to check for right angles
Slopes of the lines can be found using:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Slope of [tex]\(\overline{PA}\)[/tex]:
[tex]\[ \text{Slope}_{PA} = \frac{-4 - 6}{1 - (-5)} = \frac{-4 - 6}{1 + 5} = \frac{-10}{6} = -\frac{5}{3} \][/tex]
Slope of [tex]\(\overline{AR}\)[/tex]:
[tex]\[ \text{Slope}_{AR} = \frac{6 - 9}{-5 - 0} = \frac{-3}{-5} = \frac{3}{5} \][/tex]
Slope of [tex]\(\overline{RT}\)[/tex]:
[tex]\[ \text{Slope}_{RT} = \frac{9 - (-1)}{0 - 6} = \frac{9 + 1}{0 - 6} = \frac{10}{-6} = -\frac{5}{3} \][/tex]
Slope of [tex]\(\overline{TP}\)[/tex]:
[tex]\[ \text{Slope}_{TP} = \frac{-1 - (-4)}{6 - 1} = \frac{-1 + 4}{6 - 1} = \frac{3}{5} \][/tex]
### Step 4: Verify if the slopes of consecutive sides are perpendicular
Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
Check [tex]\(\overline{PA}\)[/tex] and [tex]\(\overline{AR}\)[/tex]:
[tex]\[ \text{Slope}_{PA} \cdot \text{Slope}_{AR} = -\frac{5}{3} \cdot \frac{3}{5} = -1 \][/tex]
Check [tex]\(\overline{AR}\)[/tex] and [tex]\(\overline{RT}\)[/tex]:
[tex]\[ \text{Slope}_{AR} \cdot \text{Slope}_{RT} = \frac{3}{5} \cdot -\frac{5}{3} = -1 \][/tex]
Check [tex]\(\overline{RT}\)[/tex] and [tex]\(\overline{TP}\)[/tex]:
[tex]\[ \text{Slope}_{RT} \cdot \text{Slope}_{TP} = -\frac{5}{3} \cdot \frac{3}{5} = -1 \][/tex]
Check [tex]\(\overline{TP}\)[/tex] and [tex]\(\overline{PA}\)[/tex]:
[tex]\[ \text{Slope}_{TP} \cdot \text{Slope}_{PA} = \frac{3}{5} \cdot -\frac{5}{3} = -1 \][/tex]
All the adjacent sides are perpendicular (i.e., slopes multiply to [tex]\(-1\)[/tex]), confirming that each angle in the quadrilateral is a right angle.
### Conclusion:
Since opposite sides are equal in length and all angles are right angles, we have proven that quadrilateral PART is a rectangle.
1. Opposite sides are equal in length (PARALLELOGRAM): This ensures that the figure is a parallelogram.
2. One angle is 90 degrees (RIGHT ANGLE): This ensures that it is indeed a rectangle.
### Step 1: Calculate the lengths of all sides using the distance formula
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let’s calculate the lengths of [tex]\(\overline{PA}\)[/tex], [tex]\(\overline{AR}\)[/tex], [tex]\(\overline{RT}\)[/tex], and [tex]\(\overline{TP}\)[/tex]:
1. Length of [tex]\(\overline{PA}\)[/tex]:
[tex]\[ PA = \sqrt{(1 - (-5))^2 + (-4 - 6)^2} = \sqrt{(1 + 5)^2 + (-4 - 6)^2} = \sqrt{6^2 + (-10)^2} = \sqrt{36 + 100} = \sqrt{136} \][/tex]
2. Length of [tex]\(\overline{AR}\)[/tex]:
[tex]\[ AR = \sqrt{(-5 - 0)^2 + (6 - 9)^2} = \sqrt{(-5)^2 + (6 - 9)^2} = \sqrt{25 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
3. Length of [tex]\(\overline{RT}\)[/tex]:
[tex]\[ RT = \sqrt{(0 - 6)^2 + (9 - (-1))^2} = \sqrt{(0 - 6)^2 + (9 + 1)^2} = \sqrt{(-6)^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \][/tex]
4. Length of [tex]\(\overline{TP}\)[/tex]:
[tex]\[ TP = \sqrt{(6 - 1)^2 + (-1 - (-4))^2} = \sqrt{(6 - 1)^2 + (-1 + 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
### Step 2: Check if opposite sides are equal
From our calculations, we can see that:
- Length of [tex]\(\overline{PA} = \sqrt{136}\)[/tex]
- Length of [tex]\(\overline{RT} = \sqrt{136}\)[/tex]
And:
- Length of [tex]\(\overline{AR} = \sqrt{34}\)[/tex]
- Length of [tex]\(\overline{TP} = \sqrt{34}\)[/tex]
Thus, opposite sides are equal:
[tex]\[ PA = RT \][/tex]
[tex]\[ AR = TP \][/tex]
### Step 3: Calculate the slopes of the sides to check for right angles
Slopes of the lines can be found using:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Slope of [tex]\(\overline{PA}\)[/tex]:
[tex]\[ \text{Slope}_{PA} = \frac{-4 - 6}{1 - (-5)} = \frac{-4 - 6}{1 + 5} = \frac{-10}{6} = -\frac{5}{3} \][/tex]
Slope of [tex]\(\overline{AR}\)[/tex]:
[tex]\[ \text{Slope}_{AR} = \frac{6 - 9}{-5 - 0} = \frac{-3}{-5} = \frac{3}{5} \][/tex]
Slope of [tex]\(\overline{RT}\)[/tex]:
[tex]\[ \text{Slope}_{RT} = \frac{9 - (-1)}{0 - 6} = \frac{9 + 1}{0 - 6} = \frac{10}{-6} = -\frac{5}{3} \][/tex]
Slope of [tex]\(\overline{TP}\)[/tex]:
[tex]\[ \text{Slope}_{TP} = \frac{-1 - (-4)}{6 - 1} = \frac{-1 + 4}{6 - 1} = \frac{3}{5} \][/tex]
### Step 4: Verify if the slopes of consecutive sides are perpendicular
Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
Check [tex]\(\overline{PA}\)[/tex] and [tex]\(\overline{AR}\)[/tex]:
[tex]\[ \text{Slope}_{PA} \cdot \text{Slope}_{AR} = -\frac{5}{3} \cdot \frac{3}{5} = -1 \][/tex]
Check [tex]\(\overline{AR}\)[/tex] and [tex]\(\overline{RT}\)[/tex]:
[tex]\[ \text{Slope}_{AR} \cdot \text{Slope}_{RT} = \frac{3}{5} \cdot -\frac{5}{3} = -1 \][/tex]
Check [tex]\(\overline{RT}\)[/tex] and [tex]\(\overline{TP}\)[/tex]:
[tex]\[ \text{Slope}_{RT} \cdot \text{Slope}_{TP} = -\frac{5}{3} \cdot \frac{3}{5} = -1 \][/tex]
Check [tex]\(\overline{TP}\)[/tex] and [tex]\(\overline{PA}\)[/tex]:
[tex]\[ \text{Slope}_{TP} \cdot \text{Slope}_{PA} = \frac{3}{5} \cdot -\frac{5}{3} = -1 \][/tex]
All the adjacent sides are perpendicular (i.e., slopes multiply to [tex]\(-1\)[/tex]), confirming that each angle in the quadrilateral is a right angle.
### Conclusion:
Since opposite sides are equal in length and all angles are right angles, we have proven that quadrilateral PART is a rectangle.
