+1439 Let $O$ be the origin. Points $P$ and $Q$ lie in the first quadrant. The slope of line segment $\overline{OP}$ is $8,$ and the slope of line segment $\overline{OQ}$ is $11.$ If $OP = OQ,$ then compute the slope of line segment $\overline{PQ}.$
Note: The point $(x,y)$ lies in the first quadrant if both $x$ and $y$ are positive.
+129850 Using the circle centered at (0,0) with a radius of 4 we can find the intersection of this circle with the lines y = 8x
And y= 11x
For a line with a slope of 8, the tangent = 8 / 1 so r = sqrt ( 1^2 + 8^2) = sqrt 65)
cos = 1/sqrt (65) sin 8/ sqrt (65)
P will have the coordinates ( 4 * 1/sqrt (65) , 4* 8 /sqrt (65)) = ( 4/sqrt (65) , 32/ sqrt (65)
For a line with a slope of 11, the tangent =11/1 so r = sqrt (1^1 + 11^2) = sqrt (122)
cos = 1/ sqrt (122) sin = 11/sqrt (122)
Q will have coordinates ( 4 * 1 /sqrt (122) , 4 * 11/sqrt (122) ) = (4/sqrt(122) , 44 /sqrt (122) )
Slope between P , Q = [ 44/sqrt (122) - 32/sqrt (65) ] / [ 4/sqrt (122) - 4/ sqrt (65) ] ≈ -.108
