Math Problem Statement

Find the distance between the two points rounding to the nearest tenth if necessary: (6, -8) and (1, 4).

Solution

The problem in the image asks to find the distance between two points (6,−8)(6, -8) and (1,4)(1, 4), rounding the result to the nearest tenth if necessary.

To solve this, we use the distance formula:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Where:

  • (x1,y1)=(6,−8)(x_1, y_1) = (6, -8)
  • (x2,y2)=(1,4)(x_2, y_2) = (1, 4)

Let's calculate the distance:

= \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$ Thus, the distance between the two points is $$13$$. No rounding is necessary, as it is already an integer. Do you have any questions about this? Would you like more details on any part of the solution? Here are 5 related questions for further practice: 1. Find the distance between $$(3, 7)$$ and $$(-1, 4)$$. 2. What is the distance between $$(-2, -3)$$ and $$(4, 6)$$? 3. Calculate the distance between $$(0, 0)$$ and $$(5, 12)$$. 4. How do you derive the distance formula from the Pythagorean theorem? 5. What would be the coordinates of the midpoint between $$(6, -8)$$ and $$(1, 4)$$? **Tip**: Always ensure you subtract coordinates in the same order (e.g., $$x_2 - x_1$$ and $$y_2 - y_1$$) to avoid sign errors.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Between Points

Formulas

Distance Formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 8-10

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