The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x - 5y + 15 = 0 ?

By section-formula co-ordinates of,

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

Substituting values we get,

$P = \Big(\dfrac{2 \times -3 + 3 \times 2}{2 + 3}, \dfrac{2 \times 6 + 3 \times 1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{-6 + 6}{5}, \dfrac{12 + 3}{5}\Big) \\[1em] = \Big(\dfrac{0}{5}, \dfrac{15}{5}\Big) \\[1em] = (0, 3).$

If P will lie on the line x - 5y + 15 = 0, it will satisfy the equation.

Substituting x = 0 and y = 3 in L.H.S. of the equation x - 5y + 15 = 0.

L.H.S. = 0 - 5 × 3 + 15

= 0 - 15 + 15

= 0.

Since, L.H.S. = R.H.S.

∴ P lies on the line x - 5y + 15 = 0.

Hence, P = (0, 3) and it lies on the line x - 5y + 15 = 0.