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# Coordinates

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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.

Apr 30, 2024

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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, $$\dfrac{19 - \sqrt{442}}{9} \approx -0.2249$$.

Apr 30, 2024