DISTANCE FORMULA

The perpendicular distance of a point from the y-axis is called x-coordinate and the perpendicular distance of a point from the x-axis is called y-coordinate. The coordinates of a point on the x-axis is in form of (x, 0) and the point on the y-axis is in the form of (0, y). The x – coordinate is called abscissa and the y – coordinate is called ordinate.

The distance formula help to find out the length of the straight line between any two points. The distance formula is derived using a Pythagoras’s Theorem.

Consider two points P(x1, y1) and Q(x2, y2) on the coordinate axis. Draw PM perpendicular to OX and PR perpendicular to QN.

clip_image002

\begin{array}{l}
{\rm{OM}} = {x_1},\,{\rm{ON}} = {x_2}\\
{\rm{PM}}\, = \,{y_1},\,{\rm{QN}} = {y_2}\\
{\rm{PR}}\, = \,{x_2} - {x_1}\\
{\rm{QR}}\, = \,{y_2} - {y_1}\\
{\rm{In }}\Delta {\rm{PQR,}}
\end{array}

By Pythagoras theorem,

{\rm{P}}{{\rm{Q}}^2} = {\rm{P}}{{\rm{R}}^2} + {\rm{Q}}{{\rm{R}}^2}

Hence, the distance between the points(x1, y1) and (x2, y2)is written as follows:

{\rm{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}

Question 1:

Find the distance between two points A(2, 4) and B(4, 5).

Solution 1: 

Use the distance formula

{\rm{AB}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}

Let \left( {{x_1},{y_1}} \right) = \left( {2,4} \right){\rm{ and }}\left( {{x_2},{y_2}} \right) = \left( {4,5} \right), substitute the values as follows:

\begin{array}{c}
{\rm{AB}} = \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {5 - 4} \right)}^2}} \\
 = \sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}} \\
 = \sqrt {4 + 1} \\
 = \sqrt 5 \\
 = 2.24
\end{array}

Therefore, the distance between two points is 2.24 units.

Question 2:

Find the distance between two points P(5, 3) and Q(9, 10).

Solution 2: 

The distance between two points is determined as follows:

Use the distance formula

{\rm{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}

Let \left( {{x_1},{y_1}} \right) = \left( {5,3} \right){\rm{ and }}\left( {{x_2},{y_2}} \right) = \left( {9,10} \right), substitute the values as follows:

\begin{array}{c}
PQ = \sqrt {{{\left( {9 - 5} \right)}^2} + {{\left( {10 - 3} \right)}^2}} \\
 = \sqrt {{{\left( 4 \right)}^2} + {{\left( 7 \right)}^2}} \\
 = \sqrt {16 + 49} \\
 = \sqrt {65} \\
 \simeq 8.06
\end{array}

Therefore, the distance between two points is 8. 06 units.

Question 3:

Find a point on x – axis which is equidistant from the points (3, 2) and (–1, 3).

Solution 3:

Let a point P(x, 0) lie on x – axis which is equidistant from the points A(3, 2) and B(–1, 3).

clip_image027

clip_image029 PA = PB

Squaring both sides

(PA)2 = (PB)2

\begin{array}{l}
 \Rightarrow {\left( {x - 3} \right)^2} + {\left( {0 - 2} \right)^2} = {\left( {x + 1} \right)^2} + {\left( {0 - 3} \right)^2}\\
 \Rightarrow {x^2} - 6x + 9 + 4 = {x^2} + 2x + 1 + 9\\
 \Rightarrow 8x = 3\\
 \Rightarrow x = \frac{3}{8}
\end{array}

clip_image029[1] The coordinates of required point are \left( {\frac{3}{8},\,0} \right).

Post Author: E-Maths