Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points in the Cartesian plane (or xy –plane), at a distance ‘d’ apart.
That is
d = PQ
By the definition of coordinates,
OM = x_{1} ON = x_{2} |
MP = y_{1} NQ = y_{2} |
Step 1 :
Now, (PR ⊥ NQ)
PR = MN
(Opposite sides of the rectangle MNRP)
Step 2 :
Find the length of MN.
MN = ON - OM
MN = x_{2 }- x_{1}
Step 3 :
Find the length of RQ.
RQ = NQ - NR
RQ = y_{2 }- y_{1}
Step 4 :
Triangle PQR is right angled at R. (PR ⊥ NQ)
By Pythagorean Theorem,
PQ^{2} = PR^{2} + RQ^{2 }
d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}
By taking positive square root,
d = √[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]
Given two points (x_{1}, y_{1}) and (x_{2}, y_{2}), the distance between these points is given by the formula
√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]
Example 1 :
Find the distance between the two points given below.
(-12, 3) and (2, 5)
Solution :
Formula for the distance between the two points is
√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]
Substitute (x_{1}, y_{1}) = (-12, 3) and (x_{2}, y_{2}) = (2, 5).
= √[(2 + 12)^{2} + (5 - 3)^{2}]
= √[14^{2} + 2^{2}]
= √[196 + 4]
= √200
= √(2 ⋅ 10 ⋅ 10)
= 10√2
So, the distance between the given points is 10√2 units.
Example 2 :
Find the distance between the two points given below.
(-2, -3) and (6, -5)
Solution :
Formula for the distance between the two points is
√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]
Substitute (x_{1}, y_{1}) = (-2, -3) and (x_{2}, y_{2}) = (6, -5).
= √[(6 + 2)^{2} + (-5 + 3)^{2}]
= √[8^{2} + (-2)^{2}]
= √[64 + 4]
= √68
= √(2 ⋅ 2 ⋅ 17)
= 2√17
So, the distance between the given points is 2√17 units.
Example 3 :
If the distance between the two points given below is 2√29, then find the value of k, given that k > 0.
(-7, 2) and (3, k)
Solution :
Distance between the above two points = 2√29
√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}] = 2√29
Substitute (x_{1}, y_{1}) = (-7, 2) and (x_{2}, y_{2}) = (3, k).
√[(3 + 7)^{2} + (k - 2)^{2}] = 2√29
√[10^{2} + (k - 2)^{2}] = 2√29
√[100 + (k - 2)^{2}] = 2√29
Square both sides.
100 + (k - 2)^{2} = (2√29)^{2}
100 + k^{2} - 2(k)(2) + 2^{2} = 2^{2}(√29)^{2}
100 + k^{2} - 4k + 4 = 4(29)
k^{2} - 4k + 104 = 116
Subtract 116 from each side.
k^{2} - 4k - 12 = 0
(k - 6)(k + 2) = 0
k - 6 = 0 or k + 2 = 0
k = 6 or k = -2
Because k > 0, we have
k = 6
Example 4 :
Find the distance between the points A and B in the xy-pane shown below.
Solution :
Identify the points A and B in the xy-plane above.
Formula for the distance between the two points is
√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]
To find the distance between the points A and B, substitute (x_{1}, y_{1}) = (2, -3) and (x_{2}, y_{2}) = (5, 5).
AB = √[(5 - 2)^{2} + (5 + 3)^{2}]
AB = √[3^{2} + 8^{2}]
AB = √(9 + 64)
AB = √73 units
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