+19 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.
+10 \(Q\) 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 \(s=\large{\frac{y_2-y_2}{x_2-x_1}=\frac{\sqrt5-3}{\sqrt5-1}=\boxed{\frac{1-\sqrt5}{2}}}\)
+129895 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
