+126 A =(10,-10) and O=(0,0) Determine the sum of all x and y coordinates of all points Q on the line y =x +6 such that angle OQA =90
+129847 I don't believe this is possible, geoNewbie
The hypotenuse will be the fixed distance from O to A = sqrt (10^2 + 10^2) = sqrt (200) =OA
Call the point that we seek ( x, x+ 6)
The distance from O to Q = sqrt [ ( x^2 + ( x + 6)^2 ] = OQ
And the distance from A to Q = sqrt [ ( x - 10) + ( x + 6 - - 10)^2 ] =
sqrt [ (x - 10)^2 + (x + 16)^2 ] = AQ
Using the Law of Cosines we have that
OA^2 = OQ^2 + AQ^2 - 2( OQ * AQ) cos (90)
The cos 90 = 0 so this boils down to
OA^2 = OQ^2 + AQ^2
200 = ( x^2 + (x + 6)^2 + ( x - 10)^2 + ( x + 16)^2
Using WolframAlpha to solve this, we get no real solutions for x
Also.....playing around with this triangle in Geogebra, I couldn't find any angles of OQA even approaching 90°
In fact about the largest angle I could find is ≈ 38.xxx °
I wonder if you might not mean angle QOA is the right angle ????
+506 I believe you mean y = -x + 6, since this is worded exactly like an alcumus problem I saw a few days ago.
Anyway, the solution I did was:
The slope of the line from the origin to the point Q must be perpendicular to the slope of the line from (10, -10) to the point Q.
The equation for that is: \(\frac{x}{y} = - \frac{-10-y}{10-x}\)
Replace y with -x + 6 to solve for x:
\(\frac{x}{-x+6} = -\frac{-10-(-x+6)}{10-x}\\\frac{x}{-x+6} = \frac{16-x}{10-x} \\ 10x-x^2=x^2-22x+96\\ 2x^2-32x+96=0\\x^2-16x+48=0\\(x-4)(x-12)=0\\x=4, x=12\)
The y values corresponding to these x values are \(y = 2, y = -6\), respectively.
Therefore, the answer to this question is \(4+12+2-6=\boxed{12}\)
https://www.desmos.com/calculator/efz9crqvir (this will help you visualize it)
+126 Yes, I am sorry I had it wrong and I think I wasted everyone's time. I was struggling with it and thought I was missing something, and now I know what it was the "-" sign.
