1d距離程式既問題

distance between 2 points 兩點的長度
= √[(difference between y-coordinates)²+ (difference between x-coordinates)²]


計算兩點的長度
12.M(a,2a),N(4a,-2a) in which a > 0

distance between MN
= √{[(2a-(- 2a)]²+ (a-4a)²}
= √[(4a)² + (-3a)²]
= √(25a²)
= 5a

Determine whether the three given points in each of the following are collinear

19.P(-3,6),Q(0,0)and R(6,-3)
distance between PR = √{[(6-(-3)]² + (-3- 6)²} = √(162) = 9√(2)
distance between PQ = √[(6-0)² + (-3- 0)²] = √(45) = 3√(5)
distance between QR = √{[(0-(-3)]² + (0- 6)²] = √(45) = 3√(5)

distance PQ + distance QR
= 3√(5) + 3√(5)
= 6√(5)

distance between PR ≠ distance PQ + distance QR
therefore, P(-3,6),Q(0,0)and R(6,-3) are not collinear


20.X(2,-1),Y(4,1)and Z(7,4)
distance between XZ = √[(-1- 4)² + (2- 7)²] = √(50) = 5√(2)
distance between XY = √[(-1-1)² + (2- 4)²] = √(8) = 2√(2)
distance between YZ = √[(1- 4)² + (4- 7)²] = √(18) = 3 √(2)

distance between XY + distance between YZ
= 2√(2) + 3√(2)
= 5√(2)
= distance btween XZ

therefore, X(2,-1),Y(4,1)and Z(7,4) are collinear