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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 $8,$ and the slope of line segment $\overline{OQ}$ is $11.$ 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.

 Dec 31, 2023
 #1
avatar+129195 
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Using the circle centered at (0,0) with a radius  of 4  we can find the intersection of this  circle with the lines y = 8x

And  y= 11x

 

For a line  with a slope of  8, the tangent = 8 / 1   so   r = sqrt ( 1^2 + 8^2)  =  sqrt 65)

cos =  1/sqrt (65)     sin  8/ sqrt (65)

 

 

P  will  have the coordinates  (  4 * 1/sqrt (65) , 4* 8 /sqrt (65)) =  ( 4/sqrt (65) , 32/ sqrt (65)

 

For a line with  a slope of  11, the tangent  =11/1 so r  = sqrt (1^1 + 11^2)  = sqrt (122)

cos =  1/ sqrt (122)    sin = 11/sqrt (122)

 

Q will have coordinates  ( 4 * 1 /sqrt (122) , 4 * 11/sqrt (122) ) =   (4/sqrt(122) , 44 /sqrt (122) )

 

Slope  between P , Q   =    [ 44/sqrt (122) - 32/sqrt (65) ] /  [ 4/sqrt (122) - 4/ sqrt (65) ]  ≈ -.108

 

 

cool cool cool

 Jan 1, 2024

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