The line segment joining points A(2,1) and B(5,-8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x -y + K = 0, find the value of k.

Answer

To solve the problem, we need to find the value of \( K \) such that the point \( P \) lies on the line given by the equation \( 2x - y + K = 0 \). The point \( P \) is the trisection point on the line segment joining points \( A(2, 1) \) and \( B(5, -8) \).

### Step 1: Find the coordinates of point P using the section formula.
Since \( P \) is nearer to \( A \) and trisects the line segment \( AB \), we can use the section formula. The ratio of the segments \( AP \) and \( PB \) is \( 1:2 \).

Using the section formula:
\[ P\left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \]
where \( m_1 = 1 \), \( m_2 = 2 \), \( A(2, 1) \) is \( (x_1, y_1) \) and \( B(5, -8) \) is \( (x_2, y_2) \).

### Step 2: Substitute the values into the section formula.
\[ P_x = \frac{1 \cdot 5 + 2 \cdot 2}{1 + 2} = \frac{5 + 4}{3} = \frac{9}{3} = 3 \]
\[ P_y = \frac{1 \cdot (-8) + 2 \cdot 1}{1 + 2} = \frac{-8 + 2}{3} = \frac{-6}{3} = -2 \]
Thus, the coordinates of point \( P \) are \( (3, -2) \).

### Step 3: Substitute the coordinates of point P into the line equation.
Now we substitute \( P(3, -2) \) into the line equation \( 2x - y + K = 0 \):
\[ 2(3) - (-2) + K = 0 \]
\[ 6 + 2 + K = 0 \]
\[ 8 + K = 0 \]

### Step 4: Solve for K.
\[ K = -8 \]

### Conclusion
The value of \( K \) is \( -8 \).

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