Find the coordinates of the point P which is three-fourth of the way from A(3, 1) to B(-2, 5).

Coordinates of A(3, 1) and B(-2, 5).

Let P divides AB in ratio m₁ : m₂. Given P lies on AB such that,

AP = $\dfrac{3}{4}$AB = $\dfrac{3}{4}$(AP + PB)
⇒ 4AP = 3AP + 3PB
⇒ 4AP - 3AP = 3PB
⇒ AP = 3PB
⇒ AP : PB = 3 : 1.

∴ m₁ : m₂ = 3 : 1.

Let coordinates of P be (x, y). By section formula,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{(3 \times (-2) + 1 \times 3)}{3 + 1} \\[1em] = \dfrac{-6 + 3}{4} \\[1em] = -\dfrac{3}{4}.

Similarly applying section formula we get y-coordinate,

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 5 + 1 \times 1}{3 + 1} \\[1em] = \dfrac{15 + 1}{4} \\[1em] = \dfrac{16}{4} \\[1em] = 4.

∴ P = $\Big(-\dfrac{3}{4}, 4\Big).$

Hence, coordinates of P are $\Big(-\dfrac{3}{4}, 4\Big).$