SOLUTION: 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
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Question 181784: 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
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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 x-coordinate of P equals the y-coordinate of P and x-coordinate of Q equals the y-coordinate of Q. That means the segment PQ lies on the line described by:
The distance formula:
^2 + (y_1 - y_2)^2})
gives us that the distance from P to Q is:
)^2 + (8 - (-7))^2}= sqrt{2 \ast 15^2} = 15 sqrt{2})
One eighth of that distance is then
The distance from P(8,8) to the desired point, call it R(x, y) is then given by;
^2 + (8 - y)^2})
But remember that this point must lie on the line
So we can re-write the distance equation:
^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 
, meaning that we can now write:
sqrt{2} = \frac{15 sqrt{2}}{8})
Multiplying both sides by 
yields:
Collect, apply LCD of 8, and multiply by -1:
And since
And finally, the desired point is )
John
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