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# Geometry

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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.

Jul 21, 2024

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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 :)

Jul 21, 2024