Calculate the ratio in which the line segment joining A(-4, 2) and B(3, 6) is divided by the point P(x, 3). Also, find

(i) x

(ii) length of AP.

(i) Let the point P(x, 3) divide the line segment joining A(-4, 2) and B(3, 6) in the ratio m₁ : m₂.

By section-formula,

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

Substituting values we get :

$\Rightarrow 3 = \Big(\dfrac{m$1(6) + m2(2)}{m1 + m2}\Big) \\[1em] \Rightarrow 3(m1 + m2) = {6m1 + 2m2} \\[1em] \Rightarrow 3m1 + 3m2 = 6m1 + 2m2 \\[1em] \Rightarrow 3m2 - 2m2 = 6m1 - 3m1 \\[1em] \Rightarrow m2 = 3m1 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{1}{3}.

Solving for x-coordinate with the ratio $m$1 = 1 and $m2 = 3$:

$\Rightarrow x = \dfrac{(1)(3) + (3)(-4)}{1 + 3} \\[1em] = \frac{3 - 12}{4} \\[1em] = \frac{-9}{4}.$

Hence, x = $\dfrac{-9}{4}$.

(ii) Solving length of AP

By using Distance Formula

$D = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

A(-4, 2) and $P\Big(\dfrac{-9}{4}, 3\Big)$.

$AP = \sqrt{\Big(-\dfrac{9}{4} - (-4)\Big)^2 + (3 - 2)^2} \\[1em] =\sqrt{\Big(-\dfrac{9}{4} + \dfrac{16}{4}\Big)^2 + (1)^2} \\[1em] = \sqrt{\Big(\dfrac{7}{4}\Big)^2 + 1^2} \\[1em] = \sqrt{\dfrac{49}{16} + \dfrac{16}{16}} \\[1em] = \sqrt{\dfrac{65}{16}} \\[1em] = \dfrac{\sqrt{65}}{4} \text{ units}$

Hence, length of AP = $\dfrac{\sqrt{65}}{4}$ units.