midpoint formula question

sbsbsbsbsb

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Mar 9, 2010
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Ok the directions say that m is the midpoint of line AB, where the coordinates of A are given. Find coordinates of B

A(r,s) M(0,2)

How would i find b?
 
The midpoint formula is just the coordinates of the halfway point between two points on a line.

\(\displaystyle M_{x}=\frac{x_{1}+x_{2}}{2}, \;\ M_{y}=\frac{y_{1}+y_{2}}{2}\)

But, we have to find the end point given the midpoint and the other end of the line segment.

Let the coordinates of \(\displaystyle B =(x,y), \;\ A=(r,s), \;\ M=(0,2)\)

\(\displaystyle \frac{r+x}{2}=0, \;\ \frac{s+y}{2}=2\)

\(\displaystyle x=-r, \;\ y=4-s\)

B has coordinates \(\displaystyle (-r, \;\ 4-s)\)

Use some arbitrary values to check it out.
 
Hello, sbsbsbsbsb!

\(\displaystyle M\) is the midpoint of line \(\displaystyle AB\).
. . Find the coordinates of \(\displaystyle B.\)

. . \(\displaystyle A(r,s)\qquad M(0,2)\)

If you're desperate, you can make a sketch
. . and baby-talk your way through it.

Code:
                  |         A(r,s)
                  |         o
                  |         :
                  |         : s-2
                 M|(0,2)    :
        + - - - - o - - - - +
        :         |    r
  ------:---------+------------------
        :         |
        o         |
        B         |
                  |

Point \(\displaystyle M\) is exactly halfway between \(\displaystyle A\) and \(\displaystyle B.\)


\(\displaystyle \text{Going from }M\text{ to }A\text{, we move: }\:r\text{ units right }\,\text{ and }\,s-2\text{ units up.}\)


Going from \(\displaystyle M\) to \(\displaystyle B\), we must move "the exact opposite":
. . . \(\displaystyle r \text{ units }le\!ft\,\text{ and }\,s-2\text{ units }down.\)

\(\displaystyle \text{Hence: }\;\begin{Bmatrix}x &=& 0 - r &=& -r \\ \\[-3mm] y &=& 2 - (s-2) &=& 4 - s \end{Bmatrix}\)


\(\displaystyle \text{Therefore: }\;B(-r,\:4-s)\)

 
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