+346 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.
+129895 Draw a unit circle at the origin, O, and let P and Q loe pn this circle
tan OP = 4/1
cos OP = 1/sqrt(4^2 + 1^2) = 1/sqrt 17 sin OP = 4/sqrt 17
tan OQ = 5/1
cos OQ = 1/sqrt (5^2 + 1^2) = 1/sqrt 26 sin QP = 5/sqrt 26
We can write P as ( 1/sqrt 17 , 4/sqrt 17)
We can write Q as ( 1/sqrt 26 , 5/sqrt 26)
Slope PQ =
(5/sqrt 26 - 4/sqrt 17) / ( 1/sqrt 26 - 1/sqrt 17) =
(5sqrt 17- 4sqrt 26) / ( sqrt 17 - sqrt 26) =
( 5 sqrt17 -4sqrt 26) (sqrt 17 + sqrt 26) / (17 -26) =
(5*17 + sqrt 26 * sqrt 17 - 4*26) / -9
(-19 + sqrt 442) / -9 =
(sqrt 442 - 19) / 9 ≈ -0.224
