We will discuss here how to find the distance between two points in a plane using the distance formula. As, we know the coordinates of two points in a plain fix the positions of the points in the plane and also the distance between them. The distance and the coordinates of the two points are related by an algebraic relation which can be deduced as shown below.
Let M (x\(_{1}\), y\(_{1}\)) and N (x\(_{2}\), y\(_{2}\)) are the two points in the plane. OX and OY being the rectangular axes of reference. Let MN = d. Draw MP ⊥ OX, NQ ⊥ OX and MR ⊥ NQ
According to the definition of the coordinates,
OP = x\(_{1}\), MP = y\(_{1}\), OQ = x\(_{2}\), NQ = y\(_{2}\)
From geometry, MR = PQ = OQ  OP = x\(_{2}\)  x\(_{1}\), and
NR = NQ  RQ = NQ  MP = y\(_{2}\)  y\(_{1}\).
In the rightangled triangle MRN,
MN\(^{2}\) = MR\(^{2}\) + NR\(^{2}\)
or, d\(^{2}\) = (x\(_{2}\)  x\(_{1}\))\(^{2}\) + (y\(_{2}\)  y\(_{1}\))\(^{2}\)
Therefore, d = \(\sqrt{(x_{2}  x_{1})^{2} + (y_{2}  y_{1})^{2}}\)
The distance between two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) = \(\sqrt{(x_{2}  x_{1})^{2} + (y_{2}  y_{1})^{2}}\)
= \(\sqrt{(difference of xcoordinates)^{2} + (difference of ycoordinates)^{2}}\)
The above formula is known as the distance formula.
Solved example to find the distance between two points in a plane:
Find the distance between the two points (2, 3) and (1, 1).
= \(\sqrt{(1  2)^{2} + (1  3)^{2}}\)
= \(\sqrt{(3)^{2} + (4)^{2}}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5
That is 5 units.
Note:
(i) The distance between two points is always positive.
(ii) The distance of a point (x, y) from the origin (0, 0) = \(\sqrt{(x  0)^{2} + (y  0)^{2}}\) = \(\sqrt{x^{2} + y^{2}}\)
(iii) The distance formula d\(^{2}\) = (x\(_{2}\)  x\(_{1}\))\(^{2}\) + (y\(_{2}\)  y\(_{1}\))\(^{2}\) should be understood as an algebraic relation between five variables x\(_{1}\), y\(_{1}\), x\(_{2}\), y\(_{2}\) and d. Given any four of them, the fifth variable can be known.
`● Distance and Section Formulae
10th Grade Math
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