Let O be the origin. Points P and Q lie in the first quadrant. The slope of line segment OP is 2 and the slope of line segment OQ is 3 If OP=OQ then compute the slope of line segment PQ
Note: The point (x, y) lies in the first quadrant if both x and y are positive.
Construct a circle with a center at the origin and a radius of 1
We have the equation
x^2 + y^2 = 1 (1)
Let P,Q lie on this circle
OP has a slope of 2
So y/x = 2
y = 2x
Sub this into (1) for y
x^2 + (2x)^2 =1
5x^2 = 1
x^2 = 1/5
x =1/sqrt (5)
y = 2/sqrt (5)
So....P = (1/sqrt (5) / 2sqrt (5) )
OQ has a slope of 4
So
y/x = 3
y = 3x
Sub this into (1) for y
x^2 + (3x)^2 =1
10x^2 = 1
x^2 = 1/10
x = 1/ sqrt (10)
y = 3/sqrt (10)
So....Q= (1 / sqrt (10) , 3sqrt (10) )
Slope of PQ =
[ 3/sqrt (10) - 2/sqrt (5) ] / [ 1/sqrt (10) - 1/sqrt (5) ] =
[ 3/sqrt (10) - 2sqrt (2) / sqrt (10 ] / [ 1/sqrt (10) - sqrt (2)/sqrt (10 ) ] =
[ 3 - 2sqrt (2)] [1 + sqrt (2)]
____________ * ____________ =
[ 1 - sqrt (2) ] [ 1 + sqrt (2)]
3 - 2sqrt (2) + 3sqrt (2) - 4
______________________ =
1 - 2
-1 + sqrt (2)
__________ =
1 - 2
1 - sqrt (2)