Distance Between Two Points

In order to locate the position of point in a plane or two dimensions, we require a pair of coordinate axis. The distance of the point from the centre is called x-coordinate (or abscissa) and the distance of the point from  is called y-coordinate (or ordinate). The ordered pair (x,y) represents co-ordinate of the point. Consider two points \(A(x_{1},y_{1})\;and\;B(x_{2},y_{2})\) on the given coordinate axis. The distance between these points is given as:


This study can be extended to determine the distance of two points in space. Let the points \(P(x_{1},y_{1},z_{1})\) and \(Q(x_{2},y_{2},z_{2})\) be referred to a system of rectangular axes OX,OY and OZ as shown in the figure.

3-Dimensional Point

Through the points P and Q, we draw planes parallel to the rectangular co-ordinate plane such that we get a rectangular parallelepiped with PQ as the diagonal. ∠PAQ forms a right angle and therefore, using the Pythagoras theorem in triangle PAQ ,

\(PQ^2= PA^2+AQ^2\)………(1)

Also, in triangle ANQ, ∠ANQ is a right angle. Similarly, applying the Pythagoras theorem in ΔANQ we get,


From the equations 1 and 2 we have,


As co-ordinates of the points P and Q are known,

\(PA=y_{2}-y_{1}\), \(AN=x_{2}-x_{1}\) and \(NQ=z_{2}-z_{1}\)


\(PQ^2=(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2\)

\(PQ=\sqrt{(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2}\)

This formula gives us the distance between two points \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) in three dimensions.

Distance of any point \(Q(x,y,z)\) in space from origin \(O(0,0,0)\), is given by,


Let us go through some examples to understand the distance formula in three dimensions.

Let’s Work Out:

Example: Find distance between the two points given by P(6, 4, -3) and Q(2, -8, 3).

Solution: Using distance formula to find distance between the points P and Q,

\(PQ=\sqrt{((x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2 )}\)




Example: A,B,C are three points lying on the axes x,y and z respectively, and their distances from the origin are given as  respectively; then find co-ordinates of the point which is equidistant from A,B,C and O.

Solution: Let the required point be P(x, y, z).

Co-ordinates of the points A,B and C are given as (a,0,0), (0,b,0), (0,0,c)  and (0,0,0). As we know that the point P is equidistant from the given points.

Hence, PA = PB = PC = PO

Now, applying the distance formula for PO = PA, we get




\(x= a/2\)

Similarly applying the distance formula for PO = PB and PO=PC, we get \(y= \frac{b}{2}\) and \(z= \frac{c}{2}\).

Therefore co-ordinates of the point  which are equidistant from the points A,B,C and O is given by\((\frac{a}{2},\;\frac{b}{2},\;\frac{c}{2})\)..

To learn more about three-dimensional geometry please visit our website www.byjus.com or download BYJU’s -the learning app.

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