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Chapter 2: Problem 12
Plot the points \(Q(-4,3), R(5,3), S(2,-1),\) and \(T(-7,-1) .\) Draw \(\overline{QR}, \overline{R S}, \overline{S T},\) and \(\overline{T Q} .\) What kind offigure is formed? What is its area?
Short Answer
Expert verified
The figure is a parallelogram with an area of 36 square units.
Step by step solution
01
- Plot the Points
On a coordinate plane, plot each of the given points: \( Q(-4,3) \), \( R(5,3) \), \( S(2,-1) \), and \( T(-7,-1) \).
02
- Draw the Line Segments
Connect the points as specified: Draw line segments \( \overline{QR} \), \( \overline{RS\), \( \overline{ST} \), and \( \overline{TQ} \).
03
- Identify the Figure
Analyze the plotted points and connected line segments. Observe that the figure is a parallelogram: opposite sides are equal and parallel.
04
- Calculate the Lengths of the Sides
Determine the length of each side using the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
05
- Find the Length of \( \overline{QR} \)
Using the coordinates for \( Q \) and \( R \): \[ d = \sqrt{(5 - (-4))^2 + (3 - 3)^2} = \sqrt{9^2 + 0} = 9 \] The length of \( \overline{QR} \) is 9 units.
06
- Find the Length of \( \overline{RS} \)
Using the coordinates for \( R \) and \( S \): \[ d = \sqrt{(2 - 5)^2 + (-1 - 3)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5 \] The length of \( \overline{RS} \) is 5 units.
07
- Find the Length of \( \overline{ST} \)
Using the coordinates for \( S \) and \( T \): \[ d = \sqrt{(2 - (-7))^2 + (-1 - (-1))^2} = \sqrt{(9)^2 + 0} = 9 \] The length of \( \overline{ST} \) is 9 units.
08
- Find the Length of \( \overline{TQ} \)
Using the coordinates for \( T \) and \( Q \): \[ d = \sqrt{(-7 - (-4))^2 + (-1 - 3)^2} = \sqrt{(-3)^2 + (-4)^2} = 5 \] The length of \( \overline{TQ} \) is 5 units.
09
- Verify the Parallelogram
Verify that opposite sides are equal: \( \overline{QR} = \overline{ST} = 9 \) units and \( \overline{RS} = \overline{TQ} = 5 \) units. This confirms the figure is a parallelogram.
10
- Calculate the Area of the Parallelogram
Use the formula for the area of a parallelogram, \( \text{Area} = \text{base} \times \text{height} \). The base is 9 units and the height is the vertical distance between the lines y = 3 and y = -1 (which is 4 units):\[ \text{Area} = 9 \times 4 = 36 \text{ square units} \]
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Geometry
Coordinate geometry helps us study geometric figures using a coordinate plane. This section covers plotting points, drawing lines, and identifying shapes within the coordinate system. By assigning coordinates to each point, we can visually see how points align and form geometric figures.
This approach makes solving geometry problems easier and more intuitive. To practice, start by drawing the x-axis (horizontal) and y-axis (vertical) on graph paper. Plot each point by finding its x (horizontal) and y (vertical) coordinates. In the given exercise, the points Q, R, S, and T are plotted on the coordinate plane.
Distance Formula
The distance formula helps us find the length between two points in a coordinate plane. It is derived from the Pythagorean theorem. The formula is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
By plugging in the coordinates of each point pair, we can calculate the length of line segments between them. This reveals key properties of geometric figures such as length and symmetry.
For instance, in the exercise:
- Length of \(\overline{QR}\): \[ d = \sqrt{(5 - (-4))^2 + (3 - 3)^2} = 9 \ units\]
- Length of \(\overline{RS}\): \[ d = \sqrt{(2 - 5)^2 + (-1 - 3)^2} = 5 \ units\]
- Length of \(\overline{ST}\): \[ d = \sqrt{(2 - (-7))^2 + (-1 - (-1))^2} = 9 \ units\]
- Length of \(\overline{TQ}\): \[ d = \sqrt{(-7 - (-4))^2 + (-1 - 3)^2} = 5 \ units\]
Area of Parallelogram
The area of a parallelogram can be found using the formula:
\[ \text{Area} = \text{base} \times \text{height} \]
A parallelogram has opposite sides that are equal and parallel. Identifying the base and height correctly is crucial, where the base is a side of the parallelogram and the height is the perpendicular distance from the base to the opposite side.
In the exercise, with base length 9 units and height (difference in y-coordinates of parallel lines y = 3 and y = -1) as 4 units:
\[ \text{Area} = 9 \times 4 = 36 \ \text{square units} \]
Graphing
Graphing helps visualize geometric problems clearly. It involves plotting points accurately on a coordinate plane and connecting them to reveal shapes. Start by marking the x and y axes.
Then, plot each point by locating their x and y values. Join the points with line segments in the correct order to form the geometric figure.
From our exercise:
- Points Q(-4,3), R(5,3), S(2,-1), and T(-7,-1) are plotted.
- Line segments \( \overline{QR}, \overline{RS}, \overline{ST}, \overline{TQ} \) are drawn.
- Figure formed is identified as a parallelogram.
Understanding and practicing graphing will bolster your geometric problem-solving skills. It provides a clear and structured way to approach and solve problems.
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