On the Cartesian plane, the midpoint between two points A(a,b) and B(c,d) is M(m,n) . If A is moved vertically upwards 20 units and horizontally to the right 14 units, and B is moved vertically downwards 4 units and horizontally to the left 2 units, then the new midpoint between A and B is \(M'\). What is the distance between \(M\) and \(M'\) ?

tertre
Dec 3, 2017

#1**+1 **

Let M = [ ( a + c) / 2 , (b + d) / 2 ] = (m, n)

A' = (a +14 , b + 20) B' = (c - 2, d - 4)

So M' = [ ( [a + c] + 12) / 2 , ( [b + d] + 16)/ 2 ] = ( m + 6, n + 8)

So..... the distance between M and M' is

√ [ (m + 6 - m)^2 + ( n + 8 - n)^2 ] = √ [ 6^2 + 8^2] = √ [36 + 64] = √100 =

10 units

CPhill
Dec 3, 2017