# The line segment joining the point A(2, 1) and B(5, –8) is trisected at the points P and Q such that P is nearer to A.

Question:

The line segment joining the point 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.

Solution:

Let the points A(2, 1) and B(5, –8) is trisected at the points P(xy) and Q(ab).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (xy) divides the line segment joining the points (x1y1) and (x2y2) internally in the ratio : n, then the coordinates (xy) =

$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$

Therefore, using section formula, the coordinates of P are:

$(x, y)=\left(\frac{1(5)+2(2)}{1+2}, \frac{1(-8)+2(1)}{1+2}\right)$

$\Rightarrow(x, y)=\left(\frac{5+4}{3}, \frac{-8+2}{3}\right)$

$\Rightarrow(x, y)=\left(\frac{9}{3}, \frac{-6}{3}\right)$

$\Rightarrow(x, y)=(3,-2)$

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0

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

$\Rightarrow 6+2+k=0$

$\Rightarrow k=-8$

Hence,  the values of k is –8.