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

### Steps:

1. **Find the coordinates of point \(E\)** where segments \(AB\) and \(CD\) intersect:  
   The equations of the lines containing the segments \(AB\) and \(CD\) need to be determined first.

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

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

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

3. **Find the length of segment \(AE\)**:
   Using the distance formula between points \(A(-3, 2)\) and \(E\left(\frac{1}{3}, \frac{1}{3}\right)\):
   \[
   d = \sqrt{\left(\frac{1}{3} - (-3)\right)^2 + \left(\frac{1}{3} - 2\right)^2}
   \]
   Simplifying:
   \[
   d = \sqrt{\left(\frac{1}{3} + 3\right)^2 + \left(\frac{1}{3} - 2\right)^2}
   = \sqrt{\left(\frac{1}{3} + \frac{9}{3}\right)^2 + \left(\frac{1}{3} - \frac{6}{3}\right)^2}
   = \sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{-5}{3}\right)^2}
   \]
   \[
   d = \sqrt{\frac{100}{9} + \frac{25}{9}} = \sqrt{\frac{125}{9}} = \frac{\sqrt{125}}{3} = \frac{5\sqrt{5}}{3}
   \]
   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 \(A\) 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.

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.

Solve the problem of finding the length of segment AE, where AB and CD intersect at point E on a lattice grid. Learn about using the slope and distance formulas in coordinate geometry, perfect for Grades 10-12.

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