Question 347468: The triangle inequality states: PQ + QR > PR
Prove that the equality holds if and only if the point Q lies on the line segment PR when we use the usual Euclidean distance formula.
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! Without loss of generality, we can place P at (0,0), R at (a,0), and Q at (x,y)
with a>0
then the distances are
PQ: sqrt( x^2 + y^2 )
QR: sqrt( (x-a)^2 + y^2 )
PR: a
thus we want to know when
PQ + QR = PR or
sqrt( x^2 + y^2 ) + sqrt( (x-a)^2 + y^2 ) = a squaring both sides we get
x^2 + y^2 + 2*sqrt[ (x^2+y^2) * ( (x-a)^2 + y^2) ] + (x-a)^2 + y^2 = a^2
2*sqrt[ (x^2+y^2) * ( (x-a)^2 + y^2 ) ] = 2ax - 2x^2 - 2y^2
(x^2 + y^2) * ( (x-a)^2 + y^2 ) = (ax - x^2 - y^2) ^2
(x^2+y^2)^2 - 2ax(x^2+y^2) + a^2(x^2+y^2) = a^2x^2 - 2ax(x^2+y^2) + (x^2+y^2)
a^2(x^2+y^2) = a^2x^2 since a>0 we can divide it out
x^2+y^2 = x^2
y^2 = 0
y=0
now plugging this into original equation we get
|x| + |x-a| = a
now if x<0 then we get
-x + a -x = a
-2x = 0
x=0 which contradicts x<0 thus x>=0
if x>a then we get
x + x -a = a
2x = 2a
x=a which contradicts x>a
thus 0<=x<=a
thus point for PQ+QR=PR to be true, Q must be on the line segment PR
|
|
|
