Math Problem Statement

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Part 1

Find the distance

d left parenthesis Upper A comma Upper B right parenthesisd(A,B)

between points A and B.

Upper A left parenthesis 3 comma negative 6 right parenthesisA(3,−6)​;

Upper B left parenthesis negative 2 comma negative 13 right parenthesis

Solution

To find the distance d(A,B)d(A, B) between two points A(3,6)A(3, -6) and B(2,13)B(-2, -13), we use the distance formula:

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

Where A(x1,y1)=(3,6)A(x_1, y_1) = (3, -6) and B(x2,y2)=(2,13)B(x_2, y_2) = (-2, -13).

Now, substituting the values into the formula:

d(A,B)=((2)(3))2+((13)(6))2d(A, B) = \sqrt{((-2) - (3))^2 + ((-13) - (-6))^2} d(A,B)=(5)2+(7)2d(A, B) = \sqrt{(-5)^2 + (-7)^2} d(A,B)=25+49d(A, B) = \sqrt{25 + 49} d(A,B)=74d(A, B) = \sqrt{74}

Thus, the distance between points A and B is:

d(A,B)=748.6d(A, B) = \sqrt{74} \approx 8.6

Let me know if you need more details or have any questions! Here are five related questions:

  1. How is the distance formula derived from the Pythagorean theorem?
  2. What is the significance of the distance between two points in geometry?
  3. Can you find the midpoint of points A and B?
  4. How does the distance formula change in three-dimensional space?
  5. What are some practical applications of the distance formula?

Tip: When working with square roots, simplifying them when possible helps with exact answers.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Distance Between Two Points

Formulas

Distance formula: d(A,B) = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 8-10