DISTANCE BETWEEN TWO POINTS
Let P(x1, y1) and Q(x2, y2) 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 = x1 ON = x2 |
MP = y1 NQ = y2 |
Step 1 :
Now, (PR ⊥ NQ)
PR = MN
(Opposite sides of the rectangle MNRP)
Step 2 :
Find the length of MN.
MN = ON - OM
MN = x2 - x1
Step 3 :
Find the length of RQ.
RQ = NQ - NR
RQ = y2 - y1
Step 4 :
Triangle PQR is right angled at R. (PR ⊥ NQ)
By Pythagorean Theorem,
PQ2 = PR2 + RQ2
d2 = (x2 - x1)2 + (y2 - y1)2
By taking positive square root,
d = √[(x2 - x1)2 + (y2 - y1)2]
Given two points (x1, y1) and (x2, y2), the distance between these points is given by the formula
√[(x2 - x1)2 + (y2 - y1)2]
Examples
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
√[(x2 - x1)2 + (y2 - y1)2]
Substitute (x1, y1) = (-12, 3) and (x2, y2) = (2, 5).
= √[(2 + 12)2 + (5 - 3)2]
= √[142 + 22]
= √[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
√[(x2 - x1)2 + (y2 - y1)2]
Substitute (x1, y1) = (-2, -3) and (x2, y2) = (6, -5).
= √[(6 + 2)2 + (-5 + 3)2]
= √[82 + (-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
√[(x2 - x1)2 + (y2 - y1)2] = 2√29
Substitute (x1, y1) = (-7, 2) and (x2, y2) = (3, k).
√[(3 + 7)2 + (k - 2)2] = 2√29
√[102 + (k - 2)2] = 2√29
√[100 + (k - 2)2] = 2√29
Square both sides.
100 + (k - 2)2 = (2√29)2
100 + k2 - 2(k)(2) + 22 = 22(√29)2
100 + k2 - 4k + 4 = 4(29)
k2 - 4k + 104 = 116
Subtract 116 from each side.
k2 - 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
√[(x2 - x1)2 + (y2 - y1)2]
To find the distance between the points A and B, substitute (x1, y1) = (2, -3) and (x2, y2) = (5, 5).
AB = √[(5 - 2)2 + (5 + 3)2]
AB = √[32 + 82]
AB = √(9 + 64)
AB = √73 units
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