Distance Formula

The distance \(d\) between the two points \(P\) and \(Q\) denoted by the symbols

$$P = \left( {{x_1},{y_1}} \right)$$

and

$$Q = \left( {{x_2},{y_2}} \right)$$

can be calculated by the Distance Formula

$${d = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}}} $$

It doesn’t matter which point you designate as the first \(\left( {{x_1},{y_1}} \right)\) or the second \(\left( {{x_2},{y_2}} \right)\). If you assign point P as the first point, then point Q must be the second point. Similarly, if point Q is the first point, then point P will be the second point.

Example 1: Find the distance between the points \(\left( {2, – 6} \right)\) and \(\left( { – 3,6} \right)\).

Let \(\left( {2, – 6} \right)\) be the first point.

$$\eqalign{
{x_1} &= 2 \cr
{y_1} &= – 6 \cr} $$

That means \(\left( { – 3,6} \right)\) must be the second point.

$$\eqalign{
{x_2} &= – 3 \cr
{y_2} &= 6 \cr} $$

Substituting the values into the formula, we get

$$\eqalign{
d &= \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \cr
&= \sqrt {{{\left( { – 3 – 2} \right)}^2} + {{\left[ {6 – \left( { – 6} \right)} \right]}^2}} \cr
&= \sqrt {{{\left( { – 5} \right)}^2} + {{\left( {12} \right)}^2}} \cr
&= \sqrt {25 + 144} \cr
&= \sqrt {169} \cr
d &= 13 \cr} $$

Therefore, the distance is \(13\) units.

Example 2: Using the distance formula, show that the point P\(\left( {2, – 3} \right)\) is equidistant from points Q\(\left( {1, – 1} \right)\) and R\(\left( {4, – 2} \right)\).

To prove that point \(P\) is equidistant to points \(Q\) and \(R\), we have to show that

$$d\left( {P,Q} \right) = d\left( {P,R} \right)$$

where

\(d\left( {P,Q} \right)\) is the distance between points\(P\) and \(Q\)

and

\(d\left( {P,R} \right)\) is the distance between points \(P\) and \(R\).

\(\blacksquare\) Let’s find \(d\left( {P,Q} \right)\). If point \(P\) is the first point, then point \(Q\) must be the second.

$$P = \left( {{x_1},{y_1}} \right) = \left( {2, – 3} \right)$$

$$Q = \left( {{x_2},{y_2}} \right) = \left( {1, – 1} \right)$$

Therefore,

$$\eqalign{
d\left( {P,Q} \right) &= \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \cr
&= \sqrt {{{\left( {1 – 2} \right)}^2} + {{\left[ { – 1 – \left( { – 3} \right)} \right]}^2}} \cr
&= \sqrt {{{\left( { – 1} \right)}^2} + {{\left( 2 \right)}^2}} \cr
&= \sqrt {1 + 4} \cr
d\left( {P,Q} \right) &= \sqrt 5 \cr} $$

\(\blacksquare\) Now, we find \(d\left( {P,R} \right)\). So if point \(P\) is the first point, then point \(R\) must be the second.

$$P = \left( {{x_1},{y_1}} \right) = \left( {2, – 3} \right)$$

$$R = \left( {{x_2},{y_2}} \right) = \left( {4, – 2} \right)$$

This gives us

$$\eqalign{
d\left( {P,R} \right) &= \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \cr
&= \sqrt {{{\left( {4 – 2} \right)}^2} + {{\left[ { – 2 – \left( { – 3} \right)} \right]}^2}} \cr
&= \sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}} \cr
&= \sqrt {4 + 1} \cr
d\left( {P,R} \right) &= \sqrt 5 \cr} $$

\(\Rrightarrow\) Since \(d\left( {P,Q} \right) = d\left( {P,R} \right) = \sqrt 5 \), then the point \(P\) is equidistant from points \(Q\) and \(R\).