# Let A(x_a,y_a) and B(x_b,y_b) be two points in the plane and let P(x,y) be the point that divides bar(AB) in the ratio k :1, where k>0. Show that x= (x_a+kx_b)/ (1+k) and y= (y_a+ky_b)/( 1+k)?

Nov 18, 2016

See proof below

#### Explanation:

Let's start by calculating $\vec{A B}$ and $\vec{A P}$

We start with the $x$

$\frac{\vec{A B}}{\vec{A P}} = \frac{k + 1}{k}$

$\frac{{x}_{b} - {x}_{a}}{x - {x}_{a}} = \frac{k + 1}{k}$

Multiplying and rearranging

$\left({x}_{b} - {x}_{a}\right) \left(k\right) = \left(x - {x}_{a}\right) \left(k + 1\right)$

Solving for $x$

$\left(k + 1\right) x = k {x}_{b} - k {x}_{a} + k {x}_{a} + {x}_{a}$

$\left(k + 1\right) x = {x}_{a} + k {x}_{b}$

$x = \frac{{x}_{a} + k {x}_{b}}{k + 1}$

Similarly, with the $y$

$\frac{{y}_{b} - {y}_{a}}{y - {y}_{a}} = \frac{k + 1}{k}$

$k {y}_{b} - k {y}_{a} = y \left(k + 1\right) - \left(k + 1\right) {y}_{a}$

$\left(k + 1\right) y = k {y}_{b} - k {y}_{a} + k {y}_{a} + {y}_{a}$

$y = \frac{{y}_{a} + k {y}_{b}}{k + 1}$