Let $A (x_{a}, y_{a})$ $A(x_a,y_a)$ and $B (x_{b}, y_{b})$ $B(x_b,y_b)$ be two points in the plane and let $P (x, y)$ $P(x,y)$ be the point that divides $¯ ¯¯¯¯ ¯ A B$ $bar(AB)$ in the ratio $k : 1$ $k :1$ , where $k > 0$ $k>0$ . Show that $x = \frac{x_{a} + k x_{b}}{1 + k}$ $x= (x_a+kx_b)/ (1+k)$ and $y = \frac{y_{a} + k y_{b}}{1 + k}$ $y= (y_a+ky_b)/( 1+k)$ ?

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Answer 1 · 2016-11-18T17:15:17

See proof below

Let's start by calculating $- - \to A B$ $vec(AB)$ and $- - \to A P$ $vec(AP)$

We start with the $x$ $x$

$\frac{- - \to A B}{- - \to A P} = \frac{k + 1}{k}$ $vec(AB)/vec(AP)=(k+1)/k$

$\frac{x_{b} - x_{a}}{x - x_{a}} = \frac{k + 1}{k}$ $(x_b-x_a)/(x-x_a)=(k+1)/k$

Multiplying and rearranging

$(x_{b} - x_{a}) (k) = (x - x_{a}) (k + 1)$ $(x_b-x_a)(k)=(x-x_a)(k+1)$

Solving for $x$ $x$

$(k + 1) x = k x_{b} - k x_{a} + k x_{a} + x_{a}$ $(k+1)x=kx_b-kx_a+kx_a+x_a$

$(k + 1) x = x_{a} + k x_{b}$ $(k+1)x=x_a+kx_b$

$x = \frac{x_{a} + k x_{b}}{k + 1}$ $x=(x_a+kx_b)/(k+1)$

Similarly, with the $y$ $y$

$\frac{y_{b} - y_{a}}{y - y_{a}} = \frac{k + 1}{k}$ $(y_b-y_a)/(y-y_a)=(k+1)/k$

$k y_{b} - k y_{a} = y (k + 1) - (k + 1) y_{a}$ $ky_b-ky_a=y(k+1)-(k+1)y_a$

$(k + 1) y = k y_{b} - k y_{a} + k y_{a} + y_{a}$ $(k+1)y=ky_b-ky_a+ky_a+y_a$

$y = \frac{y_{a} + k y_{b}}{k + 1}$ $y=(y_a+ky_b)/(k+1)$

Let A(xa,ya)A(x_a,y_a) and B(xb,yb)B(x_b,y_b) be two points in the plane and let P(x,y)P(x,y) be the point that divides ¯¯¯¯¯¯ABbar(AB) in the ratio k:1k :1, where k>0k>0. Show that x=xa+kxb1+kx= (x_a+kx_b)/ (1+k) and y=ya+kyb1+ky= (y_a+ky_b)/( 1+k)?