Find the distance d left parenthesis Upper P 1 comma Upper P 2 right parenthesis between the given points Upper P 1 and Upper P 2 .
      Upper P 1 equals left parenthesis 0 comma 0 right parenthesis
      Upper P 2 equals left parenthesis 3 comma 5 right parenthesis

To find the distance \( d(P_1, P_2) \) between the points \( P_1(0, 0) \) and \( P_2(3, 5) \), we use the distance formula:

\[
d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Where:
- \( P_1(x_1, y_1) = (0, 0) \)
- \( P_2(x_2, y_2) = (3, 5) \)

Substitute the coordinates into the formula:

\[
d(P_1, P_2) = \sqrt{(3 - 0)^2 + (5 - 0)^2}
\]
\[
d(P_1, P_2) = \sqrt{3^2 + 5^2}
\]
\[
d(P_1, P_2) = \sqrt{9 + 25}
\]
\[
d(P_1, P_2) = \sqrt{34}
\]

Thus, the distance between the points is \( \sqrt{34} \), which is approximately 5.83 units.

Would you like more details on any step?

Here are 5 related questions:
1. How is the distance formula derived?
2. What is the significance of squaring the differences in the formula?
3. Can the distance formula be used in three dimensions?
4. How would the distance change if the points were in negative quadrants?
5. What is the geometric interpretation of the distance between two points?

**Tip:** Always simplify the values inside the square root before finding the decimal approximation.

Find the distance d(P1, P2) between the given points P1 and P2.
P1 = (0, 0)
P2 = (3, 5)

Learn how to find the distance between the points P1(0,0) and P2(3,5) using the distance formula derived from the Pythagorean Theorem. This problem demonstrates key geometry concepts and is suitable for students in Grades 8-10.

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