Question 1032331
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Since you did not provide the coordinates of points *[tex \Large P(x_P,y_P)] and *[tex \Large Q(x_Q,y_Q)], this question can only be answered in terms of the general coordinates of these points.


This is simply an exercise in repeated application of the midpoint formulas.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{y_1 + y_2}{2}]


See diagram (note the addition of point N which is the midpoint of MQ):


*[illustration midpoint_formula.jpg]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_M\ = \frac{x_P + x_Q}{2}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_M\ = \frac{y_P + y_Q}{2}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_N\ = \frac{x_M + x_Q}{2}\ =\ \frac{\frac{x_P + x_Q}{2} + x_Q}{2}\ =\ \frac{x_P + 3x_Q}{4}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_N\ = \frac{y_M + y_Q}{2}\ =\ \frac{\frac{y_P + y_Q}{2} + y_Q}{2}\ =\ \frac{y_P + 3y_Q}{4}]


Following the same pattern you end up with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_S\ = \frac{x_P + 7x_Q}{8}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_S\ = \frac{y_P + 7y_Q}{8}] 


I'll let you verify the arithmetic for yourself.


In general, if S were some fractional distance from P to Q, say *[tex \Large \frac{r}{s}], the coordinates of S, in terms of the coordinates of P and Q would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_S\ = \left(1\ -\ \frac{r}{s}\right)x_P\ +\  \frac{r}{s}x_Q] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_S\ = \left(1\ -\ \frac{r}{s}\right)y_P\ +\  \frac{r}{s}y_Q]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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