(1) If the sum of any two distances out of d(A, B), d(B, C) and d(A, C) is equal to the third , then the three points A, B and C are collinear.
we will find d(A, B), d(B, C) and d(A, C).
Co-ordinates of A = (1,-3)
Co-ordinates of B = (2,-5)
Co-ordinates of C = (-4,7)
By distance formula d(A,B) = √[(x2-x1)2+(y2-y1)2]
d(A,B) = √[(2-1)2+(-5-(-3))2]
d(A,B) = √[(1)2+(-2)2]
d(A,B) = √(1+4)
d(A,B) = √5 ……………(i)
By distance formula d(B,C) = √[(x2-x1)2+(y2-y1)2]
d(B,C) = √[(-4-2)2+(7-(-5))2]
d(B,C) = √[(-6)2+(12)2]
d(B,C) = √(36+144)
d(B,C) = √180
d(B,C) = 6√5 ………….(ii)
By distance formula d(A,C) = √[(x2-x1)2+(y2-y1)2]
d(A,C) = √[(-4-1)2+(7-(-3))2]
d(A,C) = √[(-5)2+(10)2]
d(A,C) = √(25+100)
d(A,C) = √125
d(A,C) = 5√5………….(iii)
From (i) , (ii) and (iii)
√5 +5√5 = 6√5
d(A,C)+d(A,B) = d(B,C)
Hence Points A, B, C are collinear.
(2) If the sum of any two distances out of d(L, M), d(M, N) and d(L, N) is equal to the third , then the three points L, M and N are collinear.
we will find d(L, M), d(M, N) and d(L, N).
Co-ordinates of L = (-2, 3)
Co-ordinates of M = (1, -3)
Co-ordinates of N = (5, 4)
By distance formula d(L,M) = √[(x2-x1)2+(y2-y1)2]
d(L,M) = √[(1-(-2))2+(-3-3)2]
d(L,M) = √[32+(-6)2]
d(L,M) = √(9+36)
d(L,M) = √45
d(L,M) = 3√5……..(i)
By distance formula d(M,N) = √[(x2-x1)2+(y2-y1)2]
d(M,N) = √[(5-1)2+(4-(-3))2]
d(M,N) = √[42+72]
d(M,N) = √(16+49)
d(M,N)= √65……..(ii)
By distance formula d(L,N) = √[(x2-x1)2+(y2-y1)2]
d(L,N) = √[(5-(-2))2+(4-3)2]
d(L,N) = √[72+12]
d(L,N) = √(49+1)
d(L,M) = √50
d(L,M) =5 √2……..(iii)
Adding (i) and (iii)
d(L,M)+d(L,N) = 3√5+√50 ≠ √65
d(L,M)+d(L,N) ≠ d(M,N)
Points L,M,N are not collinear.