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 \(\angle OQA = 90^\circ.\)

Guest Jun 12, 2019

#1**+2 **

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 :

CPhill Jun 12, 2019

#2**+2 **

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

hectictar Jun 12, 2019