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. Also, find the co-ordinates of point Q.

From figure,

P and Q trisects line joining the points (2, 1) and (5, -8).

Let P = (x, y) and it divides line in ratio 1 : 2.

By section formula,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2}

Substituting values we get,

$\Rightarrow x = \dfrac{1 \times 5 + 2 \times 2}{1 + 2} \\[1em] \Rightarrow x = \dfrac{5 + 4}{3} \\[1em] \Rightarrow x = \dfrac{9}{3} = 3. \\[1em] y = \dfrac{m$1y2 + m2y1}{m1 + m2}

Substituting values we get,

$\Rightarrow y = \dfrac{1 \times -8 + 2 \times 1}{1 + 2} \\[1em] \Rightarrow y = \dfrac{-8 + 2}{3} \\[1em] \Rightarrow y = \dfrac{-6}{3} = -2.$

P = (x, y) = (3, -2).

Since, P lies on the line 2x - y + k = 0,

Substituting values we get,

⇒ 2(3) - (-2) + k = 0

⇒ 6 + 2 + k = 0

⇒ k + 8 = 0

⇒ k = -8.

Let Q = (a, b) and it divides line in ratio 2 : 1.

By section formula,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2}

Substituting values we get,

$\Rightarrow a = \dfrac{2 \times 5 + 1 \times 2}{2 + 1} \\[1em] \Rightarrow a = \dfrac{10 + 2}{3} \\[1em] \Rightarrow a = \dfrac{12}{3} = 4. \\[1em] y = \dfrac{m$1y2 + m2y1}{m1 + m2}

Substituting values we get,

$\Rightarrow b = \dfrac{2 \times -8 + 1 \times 1}{2 + 1} \\[1em] \Rightarrow y = \dfrac{-16 + 1}{3} \\[1em] \Rightarrow y = \dfrac{-15}{3} = -5.$

Q = (a, b) = (4, -5).

Hence, k = -8 and Q = (4, -5).