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Let's break down and solve the problem step by step.

We are given that the line \( L_1 \) has the equation:

\[
3x + 4y = 24
\]

This line cuts the \(x\)-axis at point \(A\) and the \(y\)-axis at point \(B\). 

### Part (a): Coordinates of A and B
To find the coordinates of \(A\) (on the \(x\)-axis) and \(B\) (on the \(y\)-axis):

- At \(A\), the line intersects the \(x\)-axis, so \(y = 0\):
  \[
  3x + 4(0) = 24 \implies 3x = 24 \implies x = 8
  \]
  Hence, the coordinates of \(A\) are \((8, 0)\).

- At \(B\), the line intersects the \(y\)-axis, so \(x = 0\):
  \[
  3(0) + 4y = 24 \implies 4y = 24 \implies y = 6
  \]
  Hence, the coordinates of \(B\) are \((0, 6)\).

### Part (b): Coordinates of M
\(M\) is the midpoint of line segment \([AB]\). The midpoint formula is given by:

\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Substituting the coordinates of \(A (8, 0)\) and \(B (0, 6)\):

\[
M = \left( \frac{8 + 0}{2}, \frac{0 + 6}{2} \right) = (4, 3)
\]

Thus, the coordinates of \(M\) are \((4, 3)\).

### Part (c): Equation of Line \( L_2 \)
We are given that line \(L_2\) passes through points \(M(4, 3)\) and \(C(0, -2)\). To find the equation of line \(L_2\), we first find the slope using the slope formula:

\[
\text{slope of } L_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{4 - 0} = \frac{5}{4}
\]

Now that we have the slope, we use the point-slope form of a line equation:

\[
y - y_1 = m(x - x_1)
\]

Using point \(M(4, 3)\):

\[
y - 3 = \frac{5}{4}(x - 4)
\]

Simplifying:

\[
y - 3 = \frac{5}{4}x - 5 \implies y = \frac{5}{4}x - 5 + 3 \implies y = \frac{5}{4}x - 2
\]

Thus, the equation of line \(L_2\) is:

\[
y = \frac{5}{4}x - 2
\]

### Part (d): Lengths of MC and AC

1. **Length of \(MC\)**:

Using the distance formula between points \(M(4, 3)\) and \(C(0, -2)\):

\[
MC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 0)^2 + (3 - (-2))^2} = \sqrt{4^2 + (3 + 2)^2} = \sqrt{16 + 25} = \sqrt{41}
\]

Thus, \(MC = \sqrt{41}\).

2. **Length of \(AC\)**:

Using the distance formula between points \(A(8, 0)\) and \(C(0, -2)\):

\[
AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(8 - 0)^2 + (0 - (-2))^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}
\]

Thus, \(AC = \sqrt{68}\).

---

Do you need further clarification on any part of the solution? Here are some related questions you can explore:

1. How do we verify the equation of a line given two points?
2. What are the coordinates of the midpoint of a segment in general?
3. How do you derive the distance formula from the Pythagorean theorem?
4. Can a line be parallel to the x- or y-axis? If so, how would the equation look?
5. How does changing one point on a line affect its slope?

**Tip:** The distance formula is derived from the Pythagorean theorem and can be used to find the straight-line distance between any two points in a plane.

The line L1 shown on the set of axes below has equation 3x + 4y = 24. L1 cuts the x-axis at A and cuts the y-axis at B. (a) Write down the coordinates of A and B. (b) M is the midpoint of the line segment [AB]. Write down the coordinates of M. (c) The line L2 passes through the point M and the point C(0, −2). Write down the equation of L2. (d) Find the length of (i) MC; (ii) AC.

Learn how to find the coordinates of points A and B, determine the midpoint, write the equation of a line passing through two points, and calculate lengths using the distance formula. This problem involves key concepts in coordinate geometry and is suitable for students in Grades 9-11.

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