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**

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

Note, it doesn't matter which one you designate as the first point \(\left( {{x_1},{y_1}} \right)\) and the second point \(\left( {{x_2},{y_2}} \right)\). If you assign point P as your first point then you must call point Q as your second point. Likewise, if point Q is your first, then point P becomes your second.

**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

*\(\left( {1, - 1} \right)\) and*

**Q***\(\left( {4, - 2} \right)\).*

**R**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 \(R\).