Math Problem Statement

Work out all these math question

Find the length of the interval PQ.

Solution

To find the length of the interval PQPQ, we can use the distance formula between two points in a coordinate plane.

Given:

  • P=(3,1)P = (3, 1)
  • Q=(7,5)Q = (7, 5)

The distance formula is:

PQ=(x2x1)2+(y2y1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substituting in the coordinates of points PP and QQ:

PQ=(73)2+(51)2PQ = \sqrt{(7 - 3)^2 + (5 - 1)^2}

Calculating each part:

  1. (73)2=42=16(7 - 3)^2 = 4^2 = 16
  2. (51)2=42=16(5 - 1)^2 = 4^2 = 16

Adding these:

PQ=16+16=32PQ = \sqrt{16 + 16} = \sqrt{32}

Now, simplifying 32\sqrt{32}:

PQ=325.7PQ = \sqrt{32} \approx 5.7

Thus, the length of PQPQ to one decimal place is approximately:

PQ=5.7PQ = 5.7

Would you like further details or have any questions on this? Here are some additional questions related to this topic:

  1. How can you apply the distance formula to find the distance between two points in 3D space?
  2. What are some practical applications of the distance formula in geometry or physics?
  3. How does the Pythagorean theorem relate to the distance formula?
  4. Can you calculate the midpoint of the line segment PQPQ given its coordinates?
  5. How would the calculation change if the points PP and QQ were rotated around the origin?

Tip: When using the distance formula, always double-check the subtraction of coordinates and squaring to avoid sign errors.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Formula

Formulas

Distance formula: PQ = √((x₂ - x₁)² + (y₂ - y₁)²)

Theorems

Pythagorean theorem (used to derive distance formula)

Suitable Grade Level

Grades 7-9