The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0. Find the value of k.

To find the value of \( k \) such that point \( P \) lies on the line defined by the equation \( 2x - y + k = 0 \), we first need to determine the coordinates of points \( P \) and \( Q \) that trisect the line segment joining the points \( (2, 1) \) and \( (5, -8) \).

Finding the Trisection Points

The coordinates of the points \( (2, 1) \) and \( (5, -8) \) can be used to find the trisection points. The formula for finding a point that divides a line segment in the ratio \( m:n \) is given by:

\((x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)\)

In this case, we want to find points \( P \) and \( Q \) that divide the segment in the ratio \( 1:2 \) and \( 2:1 \), respectively.

Calculating Point P

For point \( P \), which divides the segment in the ratio \( 1:2 \), we can set \( m = 1 \) and \( n = 2 \). The coordinates of point \( P \) can be calculated as follows:

• Coordinates of \( (x_1, y_1) = (2, 1) \)
• Coordinates of \( (x_2, y_2) = (5, -8) \)

Now substituting into the formula:

\((x_P, y_P) = \left( \frac{1 \cdot 5 + 2 \cdot 2}{1+2}, \frac{1 \cdot (-8) + 2 \cdot 1}{1+2} \right)\)

Calculating the \( x \)-coordinate:

\( x_P = \frac{5 + 4}{3} = \frac{9}{3} = 3 \)

Calculating the \( y \)-coordinate:

\( y_P = \frac{-8 + 2}{3} = \frac{-6}{3} = -2 \)

Thus, the coordinates of point \( P \) are \( (3, -2) \).

Calculating Point Q

For point \( Q \), which divides the segment in the ratio \( 2:1 \), we set \( m = 2 \) and \( n = 1 \). The coordinates of point \( Q \) can be calculated similarly:

\((x_Q, y_Q) = \left( \frac{2 \cdot 5 + 1 \cdot 2}{2+1}, \frac{2 \cdot (-8) + 1 \cdot 1}{2+1} \right)\)

Calculating the \( x \)-coordinate:

\( x_Q = \frac{10 + 2}{3} = \frac{12}{3} = 4 \)

Calculating the \( y \)-coordinate:

\( y_Q = \frac{-16 + 1}{3} = \frac{-15}{3} = -5 \)

Thus, the coordinates of point \( Q \) are \( (4, -5) \).

Finding the Value of k

Now that we have the coordinates of point \( P \) as \( (3, -2) \), we can substitute these values into the line equation \( 2x - y + k = 0 \) to find \( k \).

Substituting \( x = 3 \) and \( y = -2 \):

\( 2(3) - (-2) + k = 0 \)

Calculating this gives:

\( 6 + 2 + k = 0 \)

Thus:

\( 8 + k = 0 \)

Solving for \( k \):

\( k = -8 \)

Final Result

The value of \( k \) such that point \( P \) lies on the line \( 2x - y + k = 0 \) is \( -8 \).