The line segment joining the points A(2,1)A(2, 1) and B(5,−8)B(5, -8) is trisected at the points PP and QQ such that PP is nearer to AA. If PP also lies on the line 2x−y+k=02x - y + k = 0, find the value of kk.

Answer: The value of kk is −8-8.

Step-by-step solution

Step 1: Understand Trisection and Section Formula

When a line segment is trisected, it is divided into three equal parts. Since point PP is nearer to AA, it divides the line segment ABAB in the ratio 1:21:2. This means the distance from AA to PP is one-third of the total length of ABAB, and the distance from PP to BB is two-thirds of the total length.

Step 2: Apply Section Formula to find coordinates of P

We use the section formula to find the coordinates of point PP. Given A(x1,y1)=(2,1)A(x_1, y_1) = (2, 1) and B(x2,y2)=(5,−8)B(x_2, y_2) = (5, -8), and the ratio m:n=1:2m:n = 1:2. Substitute these values into the formula to find the xx and yy coordinates of PP.

Step 3: Calculate coordinates of P

Perform the arithmetic operations to simplify the expressions for the xx and yy coordinates of PP. This gives us the exact location of point PP on the coordinate plane.

Step 4: Substitute P's coordinates into the line equation

The problem states that point P(3,−2)P(3, -2) lies on the line 2x−y+k=02x - y + k = 0. To find the value of kk, we substitute the xx and yy coordinates of PP into the equation of the line.

Step 5: Solve for k

Simplify the equation by performing the multiplication and addition. Then, isolate kk to find its value. This gives us the final answer for the unknown constant kk.

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