# Distance Formula - How to use it

## Distance Formula - How it Works - Video

### Example 1 Example 1:

The distance formula is used to find the distance between any two points. We need to label the two points; one as point one and the other as point two. We have labeled (-3, 2) as (x1, y1) and (3, 2) as (x2, y2). Now we can use the distance formula ==> √( x2 - x1)2 + (y2 - y2)2.

Now we substitute -3 for x1; 2 for y1; 3 for x2; 2 for y2.

√( 3 - (-3))2 + (2 - 2)2

• √(3 + 3)2 + (2 - 2)2 We change the double negatives to positive (addition).

• √(6)2 + (0)2 We add and/or subtract in each parenthesis.

• 36 + 0 We square each number.

• √36 We add the two numbers.

• 6 We take the square root.

In this case, since the line is horizontal, we could always just count the spaces from each point. We get the same amount.

So our final answer is 6 units.

### Example 2 Example 2:

The distance formula is used to find the distance between any two points. We need to label the two points; one as point one and the other as point two. We have labeled (-2, 4) as (x1, y1) and (-2, -3) as (x2, y2). Now we can use the distance formula ==> √( x2 - x1)2 + (y2 - y2)2.

Now we substitute -2 for x1; 4 for y1; -2 for x2; -3 for y2.

√( -2 - (-2))2 + ((-3) - 4)2

• √(-2 + 2)2 + (-3 - 4)2 We change the double negatives to positive (addition).

• √(0)2 + (-7)2 We add and/or subtract in each parenthesis.

• 0 + 49 We square each number.

• 49 We add the two numbers.

• 7 We take the square root.

In this case, since the line is vertical, we could always just count the spaces from each point. We get the same amount.

So our final answer is 7 units.

### Example 3 Example 3:

The distance formula is used to find the distance between any two points. We need to label the two points; one as point one and the other as point two. We have labeled (-4, -3) as (x1, y1) and (4, 3) as (x2, y2). Now we can use the distance formula ==> √( x2 - x1)2 + (y2 - y2)2.

Now we substitute -4 for x1; -3 for y1; 4 for x2; 3 for y2.

√( 4 - (-4))2 + (3 - (-3))2

• √(4 + 4)2 + (3 + 3)2 We change the double negatives to positive (addition).

• √(8)2 + (6)2 We add and/or subtract in each parenthesis.

• 64 + 36 We square each number.

• 100 We add the two numbers.

• 10 We take the square root.

In this case, since the line is neither horizontal nor vertical, so we have to use the distance formula. In essence we are using the Pythagorean Theorem to the distance of the hypotenuse.

So our final answer is 10 units.

### Example 4 Example 4:

The distance formula is used to find the distance between any two points. We need to label the two points; one as point one and the other as point two. We have labeled (-2, 4) as (x1, y1) and (3, -1) as (x2, y2). Now we can use the distance formula ==> √( x2 - x1)2 + (y2 - y2)2.

Now we substitute -2 for x1; 4 for y1; 3 for x2; -1 for y2.

√( 3 - (-2))2 + ((-1) - 4)2

• √(3 + 2)2 + (-1 - 4)2 We change the double negatives to positive (addition).

• √(5)2 + (-5)2 We add and/or subtract in each parenthesis.

• 25 + 25 We square each number.

• 50 We add the two numbers.

• √25*√2 We separated the square root using square root properties.

• 52 We take the square root.

In this case, since the line is neither horizontal nor vertical, so we have to use the distance formula. In essence we are using the Pythagorean Theorem to the distance of the hypotenuse.

So our final answer is 5√2 units.