#### Question

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 2*x* - *y* + *k* = 0. Find the value of k.

#### Solution

We have two points A (2, 1) and B (5,-8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining `A(x_1,y_1)` and `B(x_2,y_2)`in the ratio m: n internally than,

`P(x,y) = ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

`p(x_1,y_1) = ((1(5) + 2(2))/(1 + 2)"," (2(1) + 1(-8))/(1 + 2))`

Therefore, co-ordinates of point P is(3,-2)

It is given that point P lies on the line whose equation is `2x - y + k = 0`

So point A will satisfy this equation.

2(4) - 0 + k = 0

So,

k = -8