Let \(A=(10,-10)\) and \(O=(0,0)\). Determine the sum of all \(x\) and \(y\)-coordinates of all points \(Q\) on the line \(y=-x+6\) such that \(\angle OQA = 90^\circ.\)
+128470 Let Q = (x, -x + 6)
So....by the Pythagorean Theorem...
[Distance from (0,0) to Q]^2 + [ Distance from (10, -10) to Q ]^2 = [Distance from (0,0) to (10, -10)]^2
[ (x )^2 + ( -x + 6)^2] + [ (10-x)^2 + (-10 - (-x + 6)*^2 ] = 200
[ (x^2) + (-x + 6)^2] + [ (10 - x)^2 + ( x - 16)^2] = 200
x^2 + x^2 - 12x + 36 + x^2 - 20x + 100 + x^2 - 32x + 256 = 200
4x^2 - 64x + 192 = 0 divide through by 4
x^2 - 16x + 48 = 0 factor
(x - 12) ( x - 4) = 0
Set each factor to 0, solve for x and we get that x = 12 or x = 4
When x = 4, y = 2
When x = 12, y = -6
So...Q = (4, 2) or Q = ( 12, - 6)
So....the sum of the coordinates = 12
Here's a graph :
+9466 Let the x-coordinate Q = a
Then the y-coordinate of Q = -a + 6
slope of AQ = slope between (10, -10) and (a, -a + 6)\(\ =\ \frac{(-a+6)-(-10)}{a-10}\ =\ \frac{16-a}{a-10}\)
slope of OQ = slope between (0, 0) and (a, -a + 6)\(\ =\ \frac{(-a+6)-(0)}{a-0}\ =\ \frac{6-a}{a}\)
in order for m∠OQA to be 90°, the slope of AQ must be the negative reciprocal of the slope of OQ . So...
\(\frac{16-a}{a-10}\ =\ -(\frac{a}{6-a})\\~\\ \frac{16-a}{a-10}\ =\ \frac{-a}{6-a}\\~\\ (16-a)(6-a)\ =\ (-a)(a-10)\\~\\ 96-22a+a^2\ =\ -a^2+10a\\~\\ 2a^2-32a+96\ =\ 0\\~\\ a^2-16a+48\ =\ 0\\~\\ (a-12)(a-4)\ =\ 0\\~\\ a=12\qquad\text{or}\qquad a=4\)
There are two points for Q that make m∠OQA = 90° . They are: (12, -6) and (4, 2)
12 + -6 = 6 and 4 + 2 = 6
6 + 6 = 12
Here's another graph: https://www.desmos.com/calculator/bh5f32ciyl 
