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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 29, 2024
 #1
avatar+129487 
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Let P and Q  lie on a unit circle

 

The equation of the line through OP  is   y= 4x

The equation of the line through OQ is y = 5x

 

To find the x coordinate of P

x^2 + (4x)^2   =1

17x^2 =1

x= 1/sqrt 17      y = 4/sqrt17

 

To find the x coordinate of Q

x^2 + (5x)^2 =1

26x^2 =1

x = 1/sqrt 26     y =5/sqrt 26

 

To  find  the slope of the line through  PQ

 

[ 5/sqrt 26 - 4/sqrt 17 ] / [ 1/sqrt 26 - 1/sqrt 17 ]  =

 

[ 5 sqrt 17 - 4sqrt26 ] / [ sqrt 17 - sqrt 26 ]  =

 

[4sqrt 26 - 5sqrt 17 ] / [ sqrt 26 -sqrt 17]

 

[ sqrt 26 + sqrt 17 ] [ 4sqrt 26 - 5sqrt 17] / 9  =

 

[ 4 *26 + 4sqrt442 -5sqrt 442 - 85] / 9  =

 

[19 -  sqrt 442 ] / 9

 

The slope of the line we want  has the negative reciprocal

 

9 / [ sqrt 442 - 19 ]  =

 

9 [ sqrt 442 + 19 ] / [ 442 - 361]

 

 [sqrt 442 + 19]  /  9  ≈ 4.448

 

cool cool cool

 May 30, 2024

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