The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1 : 2. Does the line x - 2y = 0 contain Q ?

By section-formula co-ordinates of,

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

Substituting values we get,

$Q = \Big(\dfrac{1 \times 2 + 2 \times 5}{1 + 2}, \dfrac{1 \times 2 + 2 \times -4}{1 + 2}\Big) \\[1em] = \Big(\dfrac{2 + 10}{3}, \dfrac{2 + (-8)}{3}\Big) \\[1em] = \Big(\dfrac{12}{3}, \dfrac{-6}{3}\Big) \\[1em] = (4, -2).$

If Q will lie on the line x - 2y = 0, it will satisfy the equation.

Substituting x = 4 and y = -2 in L.H.S. of the equation x - 2y = 0.

L.H.S. = 4 - 2 × (-2)

= 4 + 4

= 8.

Since, L.H.S. ≠ R.H.S.

∴ Q does not lies on the line x - 2y = 0.

Hence, Q = (4, -2) and it does not lie on the line x - 2y = 0.