Here we will discuss about distance between two points.
How to find the distance between two given points?
Or,
How to find the length of the line segment joining two given points?
(A) To find the distance of a given point from the origin:
Let OX and OYbe the rectangular Cartesian Coordinates axes on the plane of reference and the Coordinates of a point P on the plane be (x, y). to find the distance of P from the origin O. from P draw PM perpendicular on OX; then , OM = x and PM = y. Now from the right angle triangle OPM we get,
OP² = OM² + PM² = x² + y²
Therefore OP = √(x² + y²) (Since, OP is positive.)
(B) To find the distance between two points whose rectangular Cartesian coordinates are given:
Let (x₁, y₁) and (x₂, y₂) be the Cartesian coordinates of the points P and Q respectively referred to rectangular coordinate axes OX and OY. We are to find the distance between the points P and Q. Draw PM and QN perpendiculars from P and Q respectively on OX; then draw PR perpendicular from P on QN.
Clearly, OM = x₁, PM = y₁, ON = x₂ and QN = y₂.
Now, PR = MN = ON  OM = x₂ – x₁
and QR = QN  RN = QN  PM = y₂ – y₁
Therefore, from the rightangled triangle PQR we get,
PQ² = PR² + QR² = (x₂  x₁)² + ( y₂  y₁)²
Therefore, PQ = √[(x₂  x₁)² + (y₂  y₁)²] (Since, PQ is positive )∙
Examples on Distance between two Points
1. Find the distance of the point (5, 12) from the origin.
Solution:
We know that, the distance between two given points (x₁, y₁) and (x₂, y₂) is
√{(x₂  x₁)² + (y₂  y₁)²}.
The required distance of the point ( 5, 12) from the origin = the distance between the points ( 5, 12) and (0, 0)
= √{( 5  0)² + (12  0)²}
= √(25 + 144)
= √169
= 13 units.
2. Find the distance between the points ( 2, 5) and (2, 2).
Solution:
We know that, the distance between two given points (x₁, y₁) and (x₂, y₂) is
√{(x₂  x₁)² + (y₂  y₁)²}.
The required distance between the given points ( 2, 5) and (2, 2)
= √{(2 + 2)² + (2  5)²}
= √(16 + 9)
= √25
= 5 units.
● Coordinate Geometry
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