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

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1 Answer
Nov 18, 2016

See proof below

Explanation:

Let's start by calculating vec(AB)AB and vec(AP)AP

We start with the xx

vec(AB)/vec(AP)=(k+1)/kABAP=k+1k

(x_b-x_a)/(x-x_a)=(k+1)/kxbxaxxa=k+1k

Multiplying and rearranging

(x_b-x_a)(k)=(x-x_a)(k+1)(xbxa)(k)=(xxa)(k+1)

Solving for xx

(k+1)x=kx_b-kx_a+kx_a+x_a(k+1)x=kxbkxa+kxa+xa

(k+1)x=x_a+kx_b(k+1)x=xa+kxb

x=(x_a+kx_b)/(k+1)x=xa+kxbk+1

Similarly, with the yy

(y_b-y_a)/(y-y_a)=(k+1)/kybyayya=k+1k

ky_b-ky_a=y(k+1)-(k+1)y_akybkya=y(k+1)(k+1)ya

(k+1)y=ky_b-ky_a+ky_a+y_a(k+1)y=kybkya+kya+ya

y=(y_a+ky_b)/(k+1)y=ya+kybk+1