Question

Point C is located at minus, 3−3. Points D and E are each 5 units away from Point C. Where are D and E located?

Answer 1

Point C is located at \((-3, -3)\). Since points D and E are each 5 units away from Point C, we can find their coordinates using the distance formula.

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Here, we want to set \(d = 5\) for points D and E around point C. We know:

\[
\sqrt{(x - (-3))^2 + (y - (-3))^2} = 5
\]

This simplifies to:

\[
\sqrt{(x + 3)^2 + (y + 3)^2} = 5
\]

To eliminate the square root, we can square both sides:

\[
(x + 3)^2 + (y + 3)^2 = 25
\]

This equation represents a circle centered at \((-3, -3)\) with a radius of 5. To find specific points D and E that lie on this circle, we can choose values for \(x\) and solve for \(y\), or vice versa.

For example, if we choose to find points directly along the x-axis and y-axis:

1. **Finding D:** Set \(y = -3\) (same y-coordinate as C):
\[
(x + 3)^2 + 0 = 25 \quad \Rightarrow \quad (x + 3)^2 = 25 \quad \Rightarrow \quad x + 3 = 5 \quad \text{or} \quad x + 3 = -5
\]
This gives:
\[
x = 2 \quad \text{or} \quad x = -8
\]
Thus, we have two points:
- D: \((2, -3)\)
- E: \((-8, -3)\)

2. **Finding alternatives:** Now, let's find points directly above and below point C:
Set \(x = -3\) (same x-coordinate as C):
\[
0 + (y + 3)^2 = 25 \quad \Rightarrow \quad (y + 3)^2 = 25 \quad \Rightarrow \quad y + 3 = 5 \quad \text{or} \quad y + 3 = -5
\]
This gives:
\[
y = 2 \quad \text{or} \quad y = -8
\]
So additional points would be:
- D: \((-3, 2)\)
- E: \((-3, -8)\)

Therefore, the points D and E can be located at:
- \((2, -3)\) and \((-8, -3)\)
or
- \((-3, 2)\) and \((-3, -8)\)

In summary, D and E can be:
1. Points D and E: \((2, -3)\) and \((-8, -3)\)
2. Points D and E: \((-3, 2)\) and \((-3, -8)\)

There are infinitely many other combinations on the circle.