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

Let P(x, y) be the required point. Then according to question,

AP = $\dfrac{3}{4}$AB

$\therefore \dfrac{AP}{AB} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AP}{AP + PB} = \dfrac{3}{4} \\[1em] \Rightarrow 4AP = 3AP + 3PB \\[1em] \Rightarrow 4AP - 3AP = 3PB \\[1em] \Rightarrow AP = 3PB \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{3}{1} \\[1em] \therefore m$1 = 3, m2 = 1.

∴ By section formula coordinates of P are

$\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Putting the values from question in above formula we get,

$\Rightarrow \Big(\dfrac{3 \times (-2) + 1 \times 3}{3 + 1}, \dfrac{3 \times 5 + 1 \times 1}{3 + 1}\Big) \\[1em] = \Big(\dfrac{-6 + 3}{4}, \dfrac{15 + 1}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, \dfrac{16}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, 4\Big).$

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