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.

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,

\(AQ^2=AN^2+NQ^2\)……..(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\((\frac{a}{2},\;\frac{b}{2},\;\frac{c}{2})\)..

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Practise This Question

In the below figure,

___ angles are present and they are ___