To find the area of the rectangle defined by the given vertices \((0,0)\), \((2,2)\), \((9,-4)\), and \((7,-6)\), we first need to confirm that these points do indeed form a rectangle.
We can check the slopes of the lines connecting the points:
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Slope between (0,0) and (2,2): \[ \text{slope} = \frac{2 - 0}{2 - 0} = 1 \]
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Slope between (2,2) and (9,-4): \[ \text{slope} = \frac{-4 - 2}{9 - 2} = \frac{-6}{7} \approx -0.857 \]
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Slope between (9,-4) and (7,-6): \[ \text{slope} = \frac{-6 + 4}{7 - 9} = \frac{-2}{-2} = 1 \]
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Slope between (7,-6) and (0,0): \[ \text{slope} = \frac{0 + 6}{0 - 7} = \frac{6}{-7} \approx -0.857 \]
The opposite sides \((0,0)\) to \((2,2)\) and \((9,-4)\) to \((7,-6)\) are parallel lines with the same slope of \(1\). Similarly, the opposite sides \((2,2)\) to \((9,-4)\) and \((7,-6)\) to \((0,0)\) are also parallel lines with the same slope of \(-0.857\). This confirms that the points form a rectangle.
Next, we can calculate the dimensions of the rectangle.
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Length of one side: Distance between \((0,0)\) and \((2,2)\): \[ d_1 = \sqrt{(2-0)^2 + (2-0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
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Length of the other side: Distance between \((2,2)\) and \((9,-4)\): \[ d_2 = \sqrt{(9-2)^2 + (-4-2)^2} = \sqrt{7^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85} \]
Now, to find the area of the rectangle, we multiply the lengths of the two sides: \[ \text{Area} = d_1 \times d_2 = (2\sqrt{2}) \times \sqrt{85} = 2\sqrt{170} \approx 2 \times 13.038 = 26.076 \text{ units}^2 \]
Rounding this, we find the approximate area is \(25.76 \text{ units}^2\).
Thus, the answer is: 25.76 units².