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# help

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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 $$\angle OQA = 90^\circ.$$

Jun 12, 2019

#1
+111438
+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 :

Jun 12, 2019
#2
+8966
+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

Jun 12, 2019
#3
+111438
+2

I think I like your method better, hectictar  ....!!!

CPhill  Jun 12, 2019