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Let O be the origin. Points P and Q lie in the first quadrant. The slope of line segment OP is 1 and the slope of line OQ segment is 3. If OP=OQ then compute the slope of line segment PQ.

 Jun 21, 2024
 #1
avatar+10 
0

\(Q\) lies on the line \(y=3x\)
\(P \) lies on the line \(y=x\)

WLOG lets \(Q=(1,3)\), therefore \(OP=OQ=\sqrt{10}\)

Thus we can say \(P=(\sqrt{5},\sqrt5)\)

Slope between 2 points is \(s=\large{\frac{y_2-y_2}{x_2-x_1}=\frac{\sqrt5-3}{\sqrt5-1}=\boxed{\frac{1-\sqrt5}{2}}}\)

 Jun 21, 2024
 #2
avatar+129725 
+1

Let  P , Q  lie on a unit circle

 

This circle has the equation x^2 + y^2  =1

 

Equation of line containg OP  is  y =x

Equation of line  containing OQ  is  y = 3x

 

The x coordinate of P  can be  found   as

x^2 + x^2 = 1

2x^2 =1

x^2  =1/2

x = 1/sqrt 2      and  y =1/sqrt 2

 

The x coordinate of Q  can be found as

x^2 + (3x)^2  = 1

10x^2 = 1

x^2  =1/10

x= 1/sqrt 10     y = 3/sqrt 10

 

Slope  between   PQ   is

 

[ 1/sqrt 2 - 3/sqrt 10 ] / [ 1/sqrt 2 - 1/sqrt 10 ]  =

 

[ sqrt 10 - 3sqrt 2 ] / [sqrt 10 - sqrt 2]  =

 

[ sqrt 10 -3sqrt 2 ] [ sqrt 10 + sqrt 2 ] / [ 10 -2]  =

 

[ 10 -2sqrt 20 - 6 ] / [ 8]  =

 

[ 4 - 4sqrt 5 ] / 8  =

 

[ 1 - sqrt 5 ] / 2

 

cool cool cool

 Jun 22, 2024

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