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.
+128475 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 :
