+936 Let $O$ be the origin. Points $P$ and $Q$ lie in the first quadrant. The slope of line segment $\overline{OP}$ is $4,$ and the slope of line segment $\overline{OQ}$ is $5.$ 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.
+129847 Let P and Q lie on a unit circle
The equation of the line through OP is y= 4x
The equation of the line through OQ is y = 5x
To find the x coordinate of P
x^2 + (4x)^2 =1
17x^2 =1
x= 1/sqrt 17 y = 4/sqrt17
To find the x coordinate of Q
x^2 + (5x)^2 =1
26x^2 =1
x = 1/sqrt 26 y =5/sqrt 26
To find the slope of the line through PQ
[ 5/sqrt 26 - 4/sqrt 17 ] / [ 1/sqrt 26 - 1/sqrt 17 ] =
[ 5 sqrt 17 - 4sqrt26 ] / [ sqrt 17 - sqrt 26 ] =
[4sqrt 26 - 5sqrt 17 ] / [ sqrt 26 -sqrt 17]
[ sqrt 26 + sqrt 17 ] [ 4sqrt 26 - 5sqrt 17] / 9 =
[ 4 *26 + 4sqrt442 -5sqrt 442 - 85] / 9 =
[19 - sqrt 442 ] / 9
The slope of the line we want has the negative reciprocal
9 / [ sqrt 442 - 19 ] =
9 [ sqrt 442 + 19 ] / [ 442 - 361]
[sqrt 442 + 19] / 9 ≈ 4.448
