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Let's solve question 89 step by step using the section formula.

### Given Information:
1. Line segment \( AB \) is intercepted by the y-axis at \( C(0, c) \) and by the x-axis at \( D(x, 0) \).
2. \( A(-2, 6) \), \( AD = 1:4 \), and \( D \) is the midpoint of \( CB \).
3. We are to find the coordinates of \( D \), \( C \), and \( B(p, q) \).

### Step-by-Step Solution:

#### Step 1: Use Section Formula for \( D \)
- The ratio \( AD : DB = 1 : 4 \). Therefore, \( D \) divides \( AB \) in the ratio 1:4.
- We can apply the section formula:
\[
D(x, y) = \left( \frac{m_2 x_1 + m_1 x_2}{m_1 + m_2}, \frac{m_2 y_1 + m_1 y_2}{m_1 + m_2} \right)
\]
where \( A(x_1, y_1) = (-2, 6) \), \( B(x_2, y_2) = (p, q) \), and the ratio \( m_1:m_2 = 1:4 \).

Thus, \( D \)'s coordinates are given by:
\[
D = \left( \frac{4(-2) + 1p}{1+4}, \frac{4(6) + 1q}{1+4} \right) = \left( \frac{-8 + p}{5}, \frac{24 + q}{5} \right)
\]
Let \( D = (x_D, 0) \) since it lies on the x-axis. This implies:
\[
\frac{24 + q}{5} = 0 \quad \Rightarrow \quad 24 + q = 0 \quad \Rightarrow \quad q = -24
\]
So, the y-coordinate of point \( B \) is \( q = -24 \).

#### Step 2: \( D \)'s x-coordinate (On x-axis)
We now have \( B(p, -24) \). Also, the x-coordinate of \( D \) is \( \frac{-8 + p}{5} \). Since \( D \) lies on the x-axis, its y-coordinate is 0, so we already know \( D = (x_D, 0) \).

Given that \( D \) lies on the x-axis, its y-coordinate is \( 0 \). Hence:
\[
\frac{24 + q}{5} = 0 \im

In the given figure, if the line segment AB is intercepted by the y-axis and x-axis at C and D, respectively, such that AC:AD = 1:4 and D is the midpoint of CB. Find the coordinates of D, C, and B.

Learn how to find the coordinates of points on a line segment intercepted by the x-axis and y-axis using the section formula and ratio division. Suitable for students in Grades 9-10.

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