+269 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.
+9664 Suppose P = (x, y). Since the slope of OP is 4, by slope formula, we have \(\dfrac{y - 0}{x - 0} = 4\), which means \(y = 4x\).
So, P = (x, 4x) for some x. By the same argument, Q = (w, 5w) for some w.
Now, we use the condition \(OP = OQ\) to find a connection between w and x. Using distance formula to calculate OP and OQ, we have
\(\sqrt{x^2 + (4x)^2} =\sqrt{w^2 + (5w)^2}\)
\(17x^2 = 26w^2\)
\(x = \sqrt{\dfrac{26}{17}}w\)
Now, the slope of PQ is \(\dfrac{5w - 4x}{w - x} = \dfrac{5w - 4\cdot \sqrt{\frac{26}{17}}w}{w - \sqrt{\frac{26}{17}} w} = \dfrac{5 \sqrt{17} - 4 \sqrt{26}}{\sqrt{17} - \sqrt{26}}\)
Rationalizing gives \(m_{PQ} = \dfrac{(5 \sqrt{17} - 4\sqrt{26})(\sqrt{17} + \sqrt{26})}{17 - 26} = \dfrac{-19 + \sqrt{442}}{-9} = \dfrac{19 - \sqrt{442}}9\).
If you just need an approximate value, \(\dfrac{19 - \sqrt{442}}{9} \approx -0.2249\).
