(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) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{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) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{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) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{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) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(L,M) = √[(1-(-2))^{2}+(-3-3)^{2}]

d(L,M) = √[3^{2}+(-6)^{2}]

d(L,M) = √(9+36)

d(L,M) = √45

d(L,M) = 3√5……..(i)

By distance formula d(M,N) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(M,N) = √[(5-1)^{2}+(4-(-3))^{2}]

d(M,N) = √[4^{2}+7^{2}]

d(M,N) = √(16+49)

d(M,N)= √65……..(ii)

By distance formula d(L,N) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(L,N) = √[(5-(-2))^{2}+(4-3)^{2}]

d(L,N) = √[7^{2}+1^{2}]

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.