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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
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 #1
avatar+79894 
+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

 

 

cool cool cool

CPhill  Dec 3, 2017

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