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Let's break down and solve part (ii) of the problem.

We are given:

- \( A(-2, -1) \)
- \( B(4, 6) \)
- \( C(6, -3) \)

### Problem:
The point \( D \) lies on line \( AB \) such that line \( CD \) is perpendicular to line \( AB \). We are asked to calculate the x-coordinate of \( D \).

### Steps:

1. **Find the slope of \( AB \)**:
   The slope \( m_{AB} \) between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by:
   \[
   m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   Substituting the coordinates of \( A(-2, -1) \) and \( B(4, 6) \):
   \[
   m_{AB} = \frac{6 - (-1)}{4 - (-2)} = \frac{7}{6}
   \]

2. **Find the slope of \( CD \)**:
   Since \( CD \) is perpendicular to \( AB \), the slopes of the two lines satisfy:
   \[
   m_{AB} \cdot m_{CD} = -1
   \]
   Hence, the slope of \( CD \), \( m_{CD} \), is:
   \[
   m_{CD} = -\frac{1}{m_{AB}} = -\frac{6}{7}
   \]

3. **Equation of line \( AB \)**:
   The equation of a line passing through two points can be written as:
   \[
   y - y_1 = m(x - x_1)
   \]
   Using the slope \( m_{AB} = \frac{7}{6} \) and point \( A(-2, -1) \):
   \[
   y + 1 = \frac{7}{6}(x + 2)
   \]
   Simplifying:
   \[
   y = \frac{7}{6}x + \frac{7}{3} - 1 = \frac{7}{6}x + \frac{4}{3}
   \]
   So, the equation of line \( AB \) is:
   \[
   y = \frac{7}{6}x + \frac{4}{3}
   \]

4. **Equation of line \( CD \)**:
   The line \( CD \) passes through \( C(6, -3) \) and has slope \( m_{CD} = -\frac{6}{7} \), so its equation is:
   \[
   y - (-3) = -\frac{6}{7}(x - 6)
   \]
   Simplifying:
   \[
   y + 3 = -\frac{6}{7}x + \frac{36}{7}
   \]
   \[
   y = -\frac{6}{7}x + \frac{36}{7} - 3 = -\frac{6}{7}x - \frac{15}{7}
   \]
   So, the equation of line \( CD \) is:
   \[
   y = -\frac{6}{7}x - \frac{15}{7}
   \]

5. **Intersection of lines \( AB \) and \( CD \)**:
   To find the coordinates of \( D \), we need to solve the system of equations of lines \( AB \) and \( CD \).

   Set the equations equal to each other:
   \[
   \frac{7}{6}x + \frac{4}{3} = -\frac{6}{7}x - \frac{15}{7}
   \]
   Let's multiply through by 42 to eliminate the denominators:
   \[
   42\left( \frac{7}{6}x + \frac{4}{3} \right) = 42\left( -\frac{6}{7}x - \frac{15}{7} \right)
   \]
   Simplifying both sides:
   \[
   49x + 56 = -36x - 90
   \]
   Combine like terms:
   \[
   49x + 36x = -90 - 56
   \]
   \[
   85x = -146
   \]
   Solving for \( x \):
   \[
   x = \frac{-146}{85} = -\frac{146}{85} \approx -1.72
   \]

Thus, the x-coordinate of point \( D \) is approximately \( -1.72 \).

Would you like further clarification or additional details on any steps? Here are some related questions:

1. How do we verify if a triangle is isosceles using the distance formula?
2. Can you explain the process for finding the y-coordinate of \( D \)?
3. How do perpendicular lines relate in terms of slopes?
4. What are alternative methods for finding the intersection of two lines?
5. How can we determine the area of triangle \( ABC \) using its vertices?

**Tip:** When solving for the intersection of two lines, always check for computational accuracy by verifying the solution in the original equations.

The point D is the point on AB such that CD is perpendicular to AB. Calculate the x-coordinate of D.

Solve for the x-coordinate of point D where CD is perpendicular to AB, given the vertices of triangle ABC. This problem covers key analytic geometry concepts such as slopes of perpendicular lines and solving a system of linear equations. Suitable for high school students.

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