+1808 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.
+16 1. Our most obvious step is to draw the points \(O, P, Q\) and label their lines' respective gradients: \(OP_m = 4, OQ_m = 5\). We are trying to find out \(PQ_m\). Where m stands for the gradient. \([y = mx + c]\)
2. Let's label point Q(a, 5a) and P(b, 4b). Since OP = OQ, that means we can equate their sum of squares together.
3. \(b^2 + (4b)^2 = a^2 + (5a)^2\)
i) \(b^2 + 16b^2 = a^2 + 25a^2\)
ii) \(17b^2 = 26a^2\)
iii) \(b\sqrt{17} = a\sqrt{26}\)
iv) Now we can express b in terms of a and produce \(b = a\sqrt{\frac{26}{17}}\)
4. To find \(PQ_m\), we need to use our gradient formula which is \(\frac{y_2-y_1}{x_2-x_1} = \frac{4b-5a}{b-a}\)
5. for simplification we see that \(\frac{4b-5a}{b-a} = \frac{4b-4a-a}{b-a} = \frac{4(b-a)-a}{b-a} = 4 - \frac{a}{b-a}\)
6. Now we can express a in terms of b which turns \(\frac{a}{b-a} = \frac{a}{a\sqrt{\frac{26}{17}}-a} \)
7. \(\frac{a}{a(\sqrt{\frac{26}{17}}-1)} = \frac{1}{\sqrt{\frac{26}{17}}-1)},\) which we rationalize by multiplying both the numerator and denominator by\( \sqrt{\frac{26}{17}}+1\) which gives us \(\frac{\sqrt{442}+17}{9}\)
8. Therefore our gradient is \(4 - \frac{\sqrt{442}+17}{9} = \frac{19-\sqrt{442}}{9}\)
I'm pretty sure this slope is negative so I'm confident this may be the right answer. However, please do check for any calculation errors I've made :)
