Let's first determine the coordinates of the points as described:
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Point A:
- "three units to the left of the origin" means -3 on the x-axis.
- "two units below the X-axis" means -2 on the y-axis.
- Therefore, Point A is at coordinates A(-3, -2).
-
Point B:
- "five units to the right of the origin" gives us 5 on the x-axis.
- "four units above the X-axis" means +4 on the y-axis.
- Thus, Point B is at coordinates B(5, 4).
-
Point C:
- "one unit to the right of the origin" gives us 1 on the x-axis.
- No vertical position is mentioned, so we can assume it is on the x-axis (y=0).
- Therefore, Point C is at coordinates C(1, 0).
Now, we need to find a point that is 4.5 units away from Point A(-3, -2). The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting Point A's coordinates into the formula, we need the distance \(d\) to be 4.5 units. Thus, we have:
\[ 4.5 = \sqrt{(x + 3)^2 + (y + 2)^2} \]
Squaring both sides gives us:
\[ 20.25 = (x + 3)^2 + (y + 2)^2 \]
Now, we can find multiple points that satisfy this equation.
To find potential points, we can start with some easy values for \(x\) and then calculate \(y\).
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If \(x = -3\): \[ (y + 2)^2 = 20.25 \implies y + 2 = \pm 4.5 \implies y = 2.5 \text{ or } -6.5 \] Points: (-3, 2.5) and (-3, -6.5)
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If \(x = 1\): \[ (1 + 3)^2 + (y + 2)^2 = 20.25 \implies 16 + (y + 2)^2 = 20.25 \implies (y + 2)^2 = 4.25 \] \(\implies y + 2 = \pm \sqrt{4.25} \implies y = -2 + \sqrt{4.25}\) or \(-2 - \sqrt{4.25}\)
Approximating \(\sqrt{4.25} \approx 2.06\):
- \(y \approx -2 + 2.06 \approx 0.06\)
- \(y \approx -2 - 2.06 \approx -4.06\)
Point: (1, 0.06) and (1, -4.06)
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If \(x = 0\): \[ (0 + 3)^2 + (y + 2)^2 = 20.25 \rightarrow 9 + (y + 2)^2 = 20.25 \implies (y + 2)^2 = 11.25 \] \(\implies y + 2 = \pm \sqrt{11.25} \implies y = -2 + \sqrt{11.25}\) or \(-2 - \sqrt{11.25}\)
Approximating \(\sqrt{11.25} \approx 3.35\):
- \(y \approx -2 + 3.35 \approx 1.35\)
- \(y \approx -2 - 3.35 \approx -5.35\)
Point: (0, 1.35) and (0, -5.35)
These calculations give us multiple points that are 4.5 units away from point A (-3, -2).
Some of these points could be:
- (-3, 2.5)
- (-3, -6.5)
- (1, 0.06)
- (1, -4.06)
- (0, 1.35)
- (0, -5.35)
You can choose any of these points based on your specific requirement or context you have in your question.