Math Problem Statement

  1. For two points I and j on a plane surface, derive the observation for distance between them in plane coordinates. If the coordinates of point I are said to be fixed, what is the implication of this statement and reflect or show it in the observation equation for distance earlier derived. ( Hint: There is no guess attempt. Take distance ( ) i.e. )

Solution

To derive the distance between two points II and JJ on a plane surface in plane coordinates, we use the Euclidean distance formula. Let's denote the coordinates of point II as (x1,y1)(x_1, y_1) and the coordinates of point JJ as (x2,y2)(x_2, y_2).

The distance dd between points II and JJ is given by:

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

Derivation

  1. Identify Coordinates:

    • Point II: (x1,y1)(x_1, y_1)
    • Point JJ: (x2,y2)(x_2, y_2)
  2. Calculate the Differences in Coordinates:

    • Difference in x-coordinates: Δx=x2x1\Delta x = x_2 - x_1
    • Difference in y-coordinates: Δy=y2y1\Delta y = y_2 - y_1
  3. Apply the Pythagorean Theorem:

    • According to the Pythagorean theorem in a right triangle, the square of the hypotenuse (distance dd) is the sum of the squares of the other two sides (differences in coordinates).
    • Thus, d2=(Δx)2+(Δy)2d^2 = (\Delta x)^2 + (\Delta y)^2
  4. Take the Square Root:

    • d=(Δx)2+(Δy)2d = \sqrt{(\Delta x)^2 + (\Delta y)^2}
    • Substituting Δx\Delta x and Δy\Delta y: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Fixed Point II

If the coordinates of point II are fixed, it means that x1x_1 and y1y_1 are constants. This has the following implications for the observation equation:

  1. Fixed Coordinates:

    • Since x1x_1 and y1y_1 are fixed, they do not change. Only the coordinates of point JJ (x2x_2 and y2y_2) are variable.
  2. Simplified Observation Equation:

    • The observation equation remains the same, but with x1x_1 and y1y_1 as constants: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
    • This implies that for any given point JJ, the distance dd is a function of the coordinates of JJ relative to the fixed point II.

Observation Equation for Fixed Point II:

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

Here, x1x_1 and y1y_1 are fixed values, and dd is computed based on the coordinates (x2,y2)(x_2, y_2) of point JJ.

Would you like further details or have any questions about this derivation?

Here are 5 more questions you might consider asking:

  1. How do you derive the midpoint between two points in a plane?
  2. What is the formula for the slope of the line passing through two points?
  3. How do you find the equation of a line given two points?
  4. Can you explain how to calculate the area of a triangle given its vertices in plane coordinates?
  5. What is the distance formula in three-dimensional space?

Tip: Always double-check your coordinate differences before substituting into the distance formula to avoid sign errors.

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Math Problem Analysis

Mathematical Concepts

Euclidean Geometry
Coordinate Geometry

Formulas

Distance formula in Euclidean space

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 9-12