Question 1161626
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The assertion made in your post cannot be proven.  A line, being of infinite length, cannot have a midpoint -- it has no endpoints.  However, presuming you meant <i>Prove that, given a line <b>segment,</b> and its midpoint, the transformed midpoint will still be the midpoint of the transformed line.</i>, the following is a valid proof.


Let *[tex \Large (x_1,y_1)] and *[tex \Large (x_2,y_2)] be the endpoints of an arbitrary line segment.  Suppose the given line segment is translated *[tex \Large h] units in the horizontal direction and *[tex \Large k] units in the vertical.  Additionally, suppose the given line segment is dilated by a factor of *[tex \Large r].  Hence the transformed line segment would have endpoints of *[tex \Large (r(x_1\,+\,h),r(y_1\,+\,k))] and *[tex \Large (r(x_2\,+\,h),r(y_2\,+\,k))]


The *[tex \Large x]-coordinate of the midpoint of the original line is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ =\ \frac{x_1\,+\,x_2}{2}]


We need to show that the expression for the transformed midpoint coordinate, *[tex \Large r(x_m\,+\,h)], is identical to the expression for the transformed midpoint in terms of the *[tex \Large x]-coordinates of the transformed endpoints.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r(x_1\,+\,h)\ +\ r(x_2\,+\,h)}{2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{rx_1\,+\,rh\ +\ rx_2\,+\,rh}{2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{rx_1\,+\ rx_2\,+\,2rh}{2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\(\frac{x_1\,+\,x_2}{2}\)\ +\ rh]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  rx_m\ +\ rh\ =\ r(x_m\ +\ h)]


The proof for the *[tex \Large y]-coordinate of the midpoint is essentially identical.


QED			

								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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