# 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:

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

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.

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,

$AQ^2=AN^2+NQ^2$……..(2)

From the equations 1 and 2 we have,

$PQ^2=PA^2+NQ^2+AN^2$

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

$PA=y_{2}-y_{1}$, $AN=x_{2}-x_{1}$ and $NQ=z_{2}-z_{1}$

Therefore,

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

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

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,

$OQ=\sqrt{(x^2+y^2+z^2)}$

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, $PQ=\sqrt{((x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2 )}$ $PQ=\sqrt{(6-2)^2+(4-(-8))^2+(-3-3)^2}$ $PQ=\sqrt{(16+144+81)}$ PQ=15.524 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 $\sqrt{x^2+y^2+z^2}=\sqrt{(x-a)^2+y^2+z^2}$ $x^2+y^2+z^2=(x-a)^2+y^2+z^2$ $x^2=(x-a)^2$ $x= a/2$ 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$(\frac{a}{2},\;\frac{b}{2},\;\frac{c}{2})$..