Question 181784
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Find the point that is one-eighth the distance from the point P(8,8) to the point Q(-7,-7) along the segment PQ.


The first thing to notice is that the <i>x</i>-coordinate of P equals the <i>y</i>-coordinate of P and <i>x</i>-coordinate of Q equals the <i>y</i>-coordinate of Q.  That means the segment PQ lies on the line described by:


*[tex \LARGE \text{          }\math y = x]


The distance formula:


*[tex \LARGE \text{          }\math d = sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}]


gives us that the distance from P to Q is:


*[tex \LARGE \text{          }\math d = sqrt{(8 - (-7))^2 + (8 - (-7))^2}= sqrt{2 \ast 15^2} = 15 sqrt{2}]


One eighth of that distance is then



*[tex \LARGE \text{          }\math \frac{d}{8} = \frac{15 sqrt{2}}{8}]


The distance from P(8,8) to the desired point, call it R(x, y) is then given by;



*[tex \LARGE \text{          }\math d = sqrt{(8 - x)^2 + (8 - y)^2}]


But remember that this point must lie on the line 


*[tex \LARGE \text{          }\math y = x]


So we can re-write the distance equation:


*[tex \LARGE \text{          }\math d = sqrt{(8 - x)^2 + (8 - x)^2}\ \ \Rightarrow\ \ d = sqrt{2(8 - x)^2}\ \ \Rightarrow\ \ d = (8 - x)sqrt{2}]


But this distance needs to be one-eighth of the distance from P to Q which we have already established to be *[tex \Large \frac{15 sqrt{2}}{8}], meaning that we can now write:


*[tex \LARGE \text{          }\math (8 - x)sqrt{2} = \frac{15 sqrt{2}}{8}]


Multiplying both sides by *[tex \Large \frac{1}{sqrt{2}}] yields:


*[tex \LARGE \text{          }\math 8 - x = \frac{15}{8}]


Collect, apply LCD of 8, and multiply by -1:


*[tex \LARGE \text{          }\math x = \frac{49}{8}]


And since


*[tex \LARGE \text{          }\math y = x \ \ \Rightarrow\ \ y = \frac{49}{8}]


And finally, the desired point is *[tex \Large \left(\frac{49}{8},\frac{49}{8}\right)]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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