Math Problem Statement

The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length (in units) of segment AE.

Solution

The problem involves finding the length of segment AEAE in a lattice of points, with key points A(3,2)A(-3, 2), B(3,1)B(3, -1), C(1,1)C(1, 1), and D(1,1)D(-1, -1), where segments ABAB and CDCD intersect at point EE. Let's break down how to approach the problem.

Steps:

  1. Find the coordinates of point EE where segments ABAB and CDCD intersect:
    The equations of the lines containing the segments ABAB and CDCD need to be determined first.

    • Line ABAB:
      The slope between A(3,2)A(-3, 2) and B(3,1)B(3, -1) is: mAB=123(3)=36=12m_{AB} = \frac{-1 - 2}{3 - (-3)} = \frac{-3}{6} = -\frac{1}{2} Using the point-slope form with point A(3,2)A(-3, 2): y - 2 = -\frac{1}{2}(x + 3) \quad \text{(equation of line AB)} Simplifying: y=12x+12y = -\frac{1}{2}x + \frac{1}{2}

    • Line CDCD:
      The slope between C(1,1)C(1, 1) and D(1,1)D(-1, -1) is: mCD=1(1)1(1)=22=1m_{CD} = \frac{1 - (-1)}{1 - (-1)} = \frac{2}{2} = 1 Using the point-slope form with point C(1,1)C(1, 1): y - 1 = 1(x - 1) \quad \text{(equation of line CD)} Simplifying: y=xy = x

  2. Solve the system of equations: To find the coordinates of point EE, solve: y=12x+12andy=xy = -\frac{1}{2}x + \frac{1}{2} \quad \text{and} \quad y = x Substituting y=xy = x into the first equation: x=12x+12x = -\frac{1}{2}x + \frac{1}{2} Multiply both sides by 2 to eliminate the fraction: 2x=x+12x = -x + 1 3x=13x = 1 x=13x = \frac{1}{3} Substituting x=13x = \frac{1}{3} into y=xy = x: y=13y = \frac{1}{3} Therefore, E(13,13)E\left(\frac{1}{3}, \frac{1}{3}\right).

  3. Find the length of segment AEAE: Using the distance formula between points A(3,2)A(-3, 2) and E(13,13)E\left(\frac{1}{3}, \frac{1}{3}\right): d=(13(3))2+(132)2d = \sqrt{\left(\frac{1}{3} - (-3)\right)^2 + \left(\frac{1}{3} - 2\right)^2} Simplifying:

    = \sqrt{\left(\frac{1}{3} + \frac{9}{3}\right)^2 + \left(\frac{1}{3} - \frac{6}{3}\right)^2} = \sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{-5}{3}\right)^2}$$ $$d = \sqrt{\frac{100}{9} + \frac{25}{9}} = \sqrt{\frac{125}{9}} = \frac{\sqrt{125}}{3} = \frac{5\sqrt{5}}{3}$$ Thus, the length of segment $$AE$$ is $$\frac{5\sqrt{5}}{3}$$ units.

Would you like further details or clarification?

Here are 5 related questions to explore:

  1. How would the solution change if the coordinates of point AA were different?
  2. What is the general formula for the intersection of two lines in 2D space?
  3. How can you apply this approach to finding distances in 3D coordinate systems?
  4. Can you derive the equation of a line given two points in vector form?
  5. How would you solve for the length of the diagonal in a 2D lattice grid?

Tip: When solving for the intersection of two lines, always express both lines in a similar form (slope-intercept) for easier substitution.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Formula
Slope of a Line

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Line Intersection Theorem
Pythagorean Theorem

Suitable Grade Level

Grades 10-12