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To solve this problem, we need to find the distance between point \( M(-6, 4) \) and the line that contains points \( C(-1, -6) \) and \( H(8, 6) \).

### Step 1: Find the equation of the line that contains points \( C(-1, -6) \) and \( H(8, 6) \).
First, we calculate the slope of the line using the formula for slope:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substitute the coordinates of points \( C(-1, -6) \) and \( H(8, 6) \):
\[
m = \frac{6 - (-6)}{8 - (-1)} = \frac{6 + 6}{8 + 1} = \frac{12}{9} = \frac{4}{3}
\]
So, the slope of the line is \( \frac{4}{3} \).

Next, we use the point-slope form of the equation of a line, which is:
\[
y - y_1 = m(x - x_1)
\]
Using point \( C(-1, -6) \) and the slope \( \frac{4}{3} \), we substitute into the equation:
\[
y - (-6) = \frac{4}{3}(x - (-1))
\]
Simplify:
\[
y + 6 = \frac{4}{3}(x + 1)
\]
\[
y + 6 = \frac{4}{3}x + \frac{4}{3}
\]
Now, subtract 6 from both sides:
\[
y = \frac{4}{3}x + \frac{4}{3} - 6
\]
To simplify further, express 6 as a fraction:
\[
y = \frac{4}{3}x + \frac{4}{3} - \frac{18}{3} = \frac{4}{3}x - \frac{14}{3}
\]
Thus, the equation of the line is:
\[
y = \frac{4}{3}x - \frac{14}{3}
\]

### Step 2: Find the distance from point \( M(-6, 4) \) to the line.
The formula for the distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
First, rewrite the equation of the line \( y = \frac{4}{3}x - \frac{14}{3} \) in standard form:
\[
y - \frac{4}{3}x + \frac{14}{3} = 0
\]
Multiply through by 3 to eliminate fractions:
\[
3y - 4x + 14 = 0
\]
Now, substitute \( A = -4 \), \( B = 3 \), and \( C = 14 \), and the coordinates of point \( M(-6, 4) \) into the distance formula:
\[
d = \frac{|(-4)(-6) + (3)(4) + 14|}{\sqrt{(-4)^2 + 3^2}}
\]
\[
d = \frac{|24 + 12 + 14|}{\sqrt{16 + 9}} = \frac{|50|}{\sqrt{25}} = \frac{50}{5} = 10
\]
Thus, the distance from point \( M(-6, 4) \) to the line is \( 10 \) units.

Would you like further details or explanations on any part of the solution? 

Here are 5 follow-up questions to extend this problem:
1. How would the distance change if point \( M \) had different coordinates?
2. Can you find the midpoint of the line segment joining points \( C(-1, -6) \) and \( H(8, 6) \)?
3. How can you verify that the slope calculation was correct?
4. What is the perpendicular distance formula for a point to a line?
5. How would you graph the line equation \( y = \frac{4}{3}x - \frac{14}{3} \) to visualize the solution?

**Tip:** Always simplify fractions early in calculations to avoid complex arithmetic later.

Find the distance between M(-6, 4) and a line that contains C(-1, -6) and H(8, 6).

Learn how to find the distance between point M(-6, 4) and a line that contains points C(-1, -6) and H(8, 6). This problem uses key concepts like slope calculation, the point-slope form of a line, and the distance from a point to a line formula. Suitable for high school students in Grades 9-12.

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