Geometry+Algebra

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.

$\begin{array}{|rcll|} \hline -\dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \\\\ \dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \quad | \quad y_Q = 6-x_Q \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{x_A-x_Q}{y_A-(6-x_Q)} &=& 0 \quad | \quad x_A = 10,\ y_A=-10 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-(6-x_Q)} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-6+x_Q} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-16+x_Q} &=& 0 \\\\ x_Q(10-x_Q) + (6-x_Q)(-16+x_Q) &=& 0 \\ 10x_Q-x_Q^2 -96+6x_Q+16x_Q-x_Q^2 &=& 0 \\ -2x_Q^2 + 32x_Q-96 &=& 0 \quad | \quad : (-2) \\ \mathbf{x_Q^2 - 16x_Q + 48} &=& \mathbf{0} \\ x_Q &=& \dfrac{16\pm \sqrt{16^2-4\cdot 48} }{2} \\ x_Q &=& \dfrac{16\pm \sqrt{64} }{2} \\ x_Q &=& \dfrac{16\pm 8 }{2} \\ x_Q &=& 8 \pm 4 \\\\ x_Q &=& 8+4 \\ \mathbf{x_Q} &=& \mathbf{12} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-12 \\ \mathbf{y_Q} &=& \mathbf{-6} \\\\ x_Q &=& 8-4 \\ \mathbf{x_Q} &=& \mathbf{4} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-4 \\ \mathbf{y_Q} &=& \mathbf{2} \\ \hline \end{array}$

The sum of all x and y-coordinates of all points Q
$12-6+4+2 = \mathbf{12}$