Name a point that is between 50 and 60 units away from (7,-2) and state the distance between the two points.
Why did the point go to the gym? Because it wanted to become "fit-ty" units away from (7,-2)! But don't worry, it's not too far, just 50 units away. The distance between the two points is, drumroll please... 50 units!
To find a point that is between 50 and 60 units away from the coordinates (7, -2), we can use the distance formula and a bit of trial and error.
Let's assume the coordinates of the point we are looking for are (x, y).
The distance formula between two points A(x₁, y₁) and B(x₂, y₂) is given by:
Distance = sqrt((x₂ - x₁)² + (y₂ - y₁)²)
In this case, let's assume the distance is 50 units. Plugging in the given values, we have:
50 = sqrt((x - 7)² + (y + 2)²)
Squaring both sides, we get:
2500 = (x - 7)² + (y + 2)²
Similarly, assuming the distance to be 60 units, we have:
3600 = (x - 7)² + (y + 2)²
Now we can solve the system of equations formed by the two equations above.
Subtracting the first equation from the second equation, we have:
3600 - 2500 = (x - 7)² + (y + 2)² - ((x - 7)² + (y + 2)²)
1100 = 0
Since 1100 is not equal to 0, the system of equations has no solution.
Therefore, there is no point between 50 and 60 units away from (7, -2).
To find a point that is between 50 and 60 units away from (7, -2), we can use the distance formula. The formula is:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Let's assume the coordinates of the point we're looking for are (x, y). We know that the distance between (7, -2) and (x, y) should be between 50 and 60. Therefore, we have the following inequality:
50 ≤ √((x - 7)^2 + (y - (-2))^2) ≤ 60
Squaring both sides of the inequality, we get:
2500 ≤ (x - 7)^2 + (y + 2)^2 ≤ 3600
Now, let's solve this inequality step by step to find the range of possible values for (x, y).
1. Lower bound: (x - 7)^2 + (y + 2)^2 ≥ 2500
Expand the equation:
x^2 - 14x + 49 + y^2 + 4y + 4 ≥ 2500
Simplify and collect like terms:
x^2 + y^2 - 14x + 4y - 2447 ≥ 0
2. Upper bound: (x - 7)^2 + (y + 2)^2 ≤ 3600
Expand the equation:
x^2 - 14x + 49 + y^2 + 4y + 4 ≤ 3600
Simplify and collect like terms:
x^2 + y^2 - 14x + 4y - 3547 ≤ 0
Now, we have a system of inequalities:
x^2 + y^2 - 14x + 4y - 2447 ≥ 0
x^2 + y^2 - 14x + 4y - 3547 ≤ 0
By graphing or using a computer algebra system, we can find the region of intersection between these two inequalities. The points within this region will satisfy the condition of being between 50 and 60 units away from (7, -2).
Lastly, to calculate the actual distance between (7, -2) and the point (x, y) within this region, use the distance formula:
Distance = √((x - 7)^2 + (y + 2)^2)
Note that there are multiple points that can satisfy this condition, so you may find different values for (x, y) and the corresponding distance using this method.
on the vertical line x=7, you could have (7+55,-2) or (7-58,-2)
On the horizontal line y = -2, you could have (7,-2+59) or (7,-2-51)
Of course, you could pick any points on the circle
(x-7)^2 + (y+2)^2 = r^2
where r is between 50 and 60.