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Let A=(10,-10) and O=(0,0). Determine the sum of all x and y-coordinates of all points Q on the line y=-2x+6 such that angle OQA = 60 degrees.

 Jul 23, 2022
 #1
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Let the point  Q =  (x , 6 - 2x)

 

The distance from  O to A   =  sqrt (200)

 

The distance  from   O to Q =  sqrt  ( x^2 + (6 -2x)^2 )  = sqrt ( 5x^2 -24x + 36)

 

The distance from  A to Q =  sqrt [ (x -10)^2  + ( 6-2x - - 10)^2 ]  = sqrt [ ( x -10)^2+ (16 - 2x)^2] =

sqrt [ 5x^2 - 84x + 356  ]

 

 

Using the Law of Cosines

 

200 =  (5x^2 - 24x + 36) + ( 5x^2 - 84x + 356)  -2 sqrt [( 5x^2 -24x + 36)(5x^2 - 84x + 356] cos (60)

 

200 = (10x^2 - 108x + 392) -2 sqrt [ (5x^2 - 24x + 36) ( 5x^2 - 84x + 356)] cos (60)

 

This is very messy to solve  (as you might expect !!!)  so I used WolframAlpha as a solver

 

It gives two values

 

x ≈ 1.5081    so    y   ≈ -2 (1.5081) + 6 = 2.9838

 

And

 

x ≈ 9.0758   so    y ≈  -2 (9.0758) + 6  = -12.1516 

 

Here's a graph :  

 

 

 

 

cool cool cool

 Jul 24, 2022

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