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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.

 May 19, 2024
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avatar+14968 
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\(\color{black }Compute\ the\ slope\ of\ line\ segment\ \overline{PQ}.\)

 

\(Let\ \overline{ OQ}=\overline{OP} =5\\ f_c(x)=+\sqrt{5^2-x^2}\\ f_{oq}(x)=5x\\ f_{op}(x)=4x\)

\(\sqrt{25-x^2}=5x\\ 25-x^2=25x^2\\ x_q= \sqrt{\frac{25}{26}}=0.9806\\ y_q=\sqrt{\frac{25}{26}}\cdot 5=4.903\\ \sqrt{25-x^2}=4x\\ x_p=\sqrt{\frac{25}{17}}=1.2127\\ y_p=\sqrt{\frac{25}{17}}\cdot 4=4.8507\)

 

\(m_{PQ}=\frac{y_q-y_p}{x_q-x_p}=\frac{4.9029-4.8507}{0.9806-1.2127}\\ \color{blue }m_{PQ}=-0.2249\\ \color{blue }The\ slope\ of\ line\ segment\ \overline{PQ}\ is\ -0.2249.\)

laugh !

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 May 19, 2024
edited by asinus  May 19, 2024

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