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Find a Point 3/8 along the Line

Date: 12/04/2002 at 21:48:10
From: Daniel 
Subject: XY graph find a point 3/8 along the line

How does one solve this one?

Find a point 3/8 from A to B.
Given: two endpoints X,Y coordinates
Point A (-2, 7) point B (6, -5)

The midpoint formula we are using is 

                X1 + X2,   Y1 + Y2
midpt (x, y) = --------    -------
                   2          2

                -2 + 6  ,  7 + -5
midpt (x, y) =  -------    ------
                   2          2
midpt (2, 1)

To find 3/8 we used:      3(X1 + X2)     3(X1 + X2)
                          ---------  ,   ----------
                              8               8

We get 3/8 point of (3/2, 3/4)

The book says it should be (1, 5/2)

Where are we messing up?
Thanks,
Daniel


Date: 12/05/2002 at 15:11:19
From: Doctor Ian
Subject: Re: XY graph find a point 3/8 along the line

Hi Daniel,

Plot the two points, 


                          |    <- 8 ->
                  A.......-........................
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .   ^
                          |                       .   |
                          -                       .  12
                          |                       .   |
                          -                       .   v
                          |                       .
   ---|---|---|---|---|---+---|---|---|---|---|---|---|---|---
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       B
                          |
                          -
                          |
                          -
                          |

Suppose we want to make a similar triangle, but 3/8 the size of this
one. The base would be (3/8)(8), or 3; and the altitude would be
(3/8)(12), or 9/2:


                     <- 3 ->

                          |   
                  A.......-........................
                          |   .                   .
                          -   .  ^                .
                          |   .  |                .
                          -   . 9/2               .
                          |   .  |                .
                          -   .  v                .    
                          |   .                   .    
                          -   .                   .   
                          |   B'                  .    
                          -                       .  
                          |                       .    
                          -                       .    
                          |                       .
   ---|---|---|---|---|---+---|---|---|---|---|---|---|---|---
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       .
                          |                       .
                          -                       B
                          |
                          -
                          |
                          -
                          |

So the location of the point you're looking for should be 

  X   = X  + (3/8)(X - X ) = -2 + (3/8)(8) = -2 + 3 = 1
   B'    A          B   A


  Y   = Y  + (3/8)(Y - Y ) = 7 + (3/8)(-12) = 7 - 9/2 = 5/2
   B'    A          B   A


Basically, to go 3/8 of the way along the hypotenuse, you have to go
3/8 of the way along each leg, starting from the initial point. 

Does this make sense?
 
- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Triangles and Other Polygons
Middle School Triangles and Other Polygons

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