Using section formula show that the points (1,-1), (2, 1) and (4, 5) are collinear.

Answer

To show that the points \( A(1, -1) \), \( B(2, 1) \), and \( C(4, 5) \) are collinear using the section formula, we will follow these steps:

### Step 1: Understand the Section Formula
The section formula states that if a point \( P(x, y) \) divides the line segment joining two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then the coordinates of point \( P \) can be calculated as:
\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]

### Step 2: Assume the Points are Collinear
Assume that points \( A \), \( B \), and \( C \) are collinear. This means point \( B \) divides the line segment \( AC \) in some ratio \( m:1 \).

### Step 3: Set Up the Coordinates
Let:
- \( A(1, -1) \) with coordinates \( (x_1, y_1) = (1, -1) \)
- \( B(2, 1) \) with coordinates \( (x_2, y_2) = (2, 1) \)
- \( C(4, 5) \) with coordinates \( (x_3, y_3) = (4, 5) \)

### Step 4: Apply the Section Formula
Using the section formula, we can express the coordinates of point \( B \) as:
\[ B\left( \frac{m \cdot 4 + 1 \cdot 1}{m+1}, \frac{m \cdot 5 + 1 \cdot (-1)}{m+1} \right) \]
This gives us two equations:
1. \( \frac{4m + 1}{m + 1} = 2 \)
2. \( \frac{5m - 1}{m + 1} = 1 \)

### Step 5: Solve the First Equation
From the first equation:
\[ \frac{4m + 1}{m + 1} = 2 \]
Cross-multiplying gives:
\[ 4m + 1 = 2(m + 1) \]
Expanding the right side:
\[ 4m + 1 = 2m + 2 \]
Rearranging terms:
\[ 4m - 2m = 2 - 1 \]
This simplifies to:
\[ 2m = 1 \quad \Rightarrow \quad m = \frac{1}{2} \]

### Step 6: Solve the Second Equation
Now, solve the second equation:
\[ \frac{5m - 1}{m + 1} = 1 \]
Cross-multiplying gives:
\[ 5m - 1 = m + 1 \]
Rearranging terms:
\[ 5m - m = 1 + 1 \]
This simplifies to:
\[ 4m = 2 \quad \Rightarrow \quad m = \frac{1}{2} \]

### Step 7: Conclusion
Since both equations yield the same value of \( m \), which is \( \frac{1}{2} \), it confirms that point \( B \) divides the segment \( AC \) in the ratio \( \frac{1}{2}:1 \). Therefore, the points \( A(1, -1) \), \( B(2, 1) \), and \( C(4, 5) \) are collinear.