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Distance Between Two Points on a Graph Study Guide (page 2)

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Updated on Oct 3, 2011

Example

Find the distance between (–2,4) and (3,16).

We will call (–2,4) point 1 and (3, 16) point 2. Substitute these values into the distance formula: D = √(x2x1)2 + (y2y1)2. Because (–2,4) is the first point, x1 is –2 and y1 is 4. The value of x2 is 3 and the value of y2 is 16:

D = √3 – (–2))2 + (16 – 4)2

Remember the order of operations: parentheses come before exponents, so perform the subtraction first:

D = √(5)2 + (12)2

Exponents come before addition, so square 5 and 12 next:

D = √25 + 144

D = √169

D = 13 units

The distance between (–2,4) and (3,16) is 13 units.

Radical Distances

Sometimes, the distance between two points is not a whole number, and we are left with a radical. Let's find the distance between (1,2) and (7,12). Enter the x and y values of each point into the distance formula:

D = √(7 – 1)2 + (12 – 1)2

D = √(6)2 + (10)2

D = √36 + 100

D = √139

The number 136 is not a perfect square. When the distance formula leaves you with a radical like this, the only way to simplify it is to factor out a perfect square. Check to see if the radicand is divisible by 4, 9, 25, or some other square. The number 136 is divisible by 4, which means that √136 is divisible by √4: √136 = √434, because we can factor a radicand into two radicands in the same way that we factor a whole number into two whole numbers. The square root of 4 is 2, so √434 is equal to 2√34. The distance between (1,2) and (7,12) is 2√34 units.

Let's look at one last example: the distance between (2,1) and (–2,10). Enter the x and y values of each point into the distance formula:

D = √((–2)– 2)2 + (10 – 1)2

D = √(–4)2 + (9)2

D = √16 + 81

D = √97

The number 97 is not divisible by any whole number perfect square (4, 25, 36, 49, 64, or 81), so √97 cannot be simplified. The distance between (2,1) and (–2,10) is √97 units.

Find practice problems and solutions for these concepts at Distance Between Two Points on a Graph Practice Questions.

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