Math Problem Statement

Calculate the distance between two points on a Cartesian plane using the given values.

Solution

The image you uploaded shows the equation for calculating the distance between two points on a Cartesian plane. Specifically, it is using the distance formula: r=(x2x1)2+(y2y1)2r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

In your equation:

  • x1=3.5x_1 = -3.5
  • y1=0y_1 = 0 (since there is no y-value with -3.5, it is assumed to be on the x-axis)
  • x2=0x_2 = 0 (since there is no x-value with 2.5, it is assumed to be on the y-axis)
  • y2=2.5y_2 = 2.5

The formula provided calculates the distance as follows: r=(3.50)2+(02.5)2r = \sqrt{(-3.5 - 0)^2 + (0 - 2.5)^2} r=(3.5)2+(2.5)2r = \sqrt{(-3.5)^2 + (-2.5)^2} r=12.25+6.25r = \sqrt{12.25 + 6.25} r=18.5r = \sqrt{18.5}

Thus, the distance rr is 18.5\sqrt{18.5}.

Do you need the numerical value of this square root, or is there anything else you'd like to know? Here are some further questions you might consider:

  1. What are the coordinates of the midpoint between these two points?
  2. How does the distance change if one of the points moves?
  3. What if this calculation were in three dimensions?
  4. Would you like to calculate the distance using a different pair of points?
  5. How can this concept be applied to real-life scenarios?

Tip: When using the distance formula, always check the signs (+ or -) of your coordinates carefully to avoid errors in your final calculation.

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Math Problem Analysis

Mathematical Concepts

Geometry
Cartesian Coordinate System

Formulas

Distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Pythagorean Theorem

Suitable Grade Level

High School