# Difference between revisions of "Working With Ordered Pairs"

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(distance numbers are all listed as 1, instead of 1, 2 3) |
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:<math>d_{1}=\sqrt{(5-2)^2 + (7-1)^2} = \sqrt{9+36} = \sqrt{45}</math> | :<math>d_{1}=\sqrt{(5-2)^2 + (7-1)^2} = \sqrt{9+36} = \sqrt{45}</math> | ||

Find the distance between line 3 and line 2. | Find the distance between line 3 and line 2. | ||

− | :<math>d_{ | + | :<math>d_{2}=\sqrt{(4-2)^2 + (0-1)^2} = \sqrt{4+1} = \sqrt{5}</math> |

Find the distance between line 3 and line 1. | Find the distance between line 3 and line 1. | ||

− | :<math>d_{ | + | :<math>d_{3}=\sqrt{(5-4)^2 + (7-0)^2} = \sqrt{1+49} = \sqrt{50}</math> |

Now to find out if it works, lets plug in our values into the [[Function_Definitions#Pythagorean_Theorem|Pythagorean Theorem]]: | Now to find out if it works, lets plug in our values into the [[Function_Definitions#Pythagorean_Theorem|Pythagorean Theorem]]: | ||

:<math>(d_{1})^2+(d_{2})^2 = (d_{3})^2</math> | :<math>(d_{1})^2+(d_{2})^2 = (d_{3})^2</math> |

## Latest revision as of 11:51, 20 April 2021

## Right Triangle Verification

Now, when faced with a set of three coordinates or ordered pairs, these coordinates must add up to **a ^{2}+b^{2}=c^{2}** which is the Pythagorean Theorem. Then simply find the distance between each set of coordinates. Once you have the distance between each set of coordinates, see if the Pythagorean Theorem applies to the set of coordinates, if so, then it is indeed a right triangle.

Example: Find the distance between line 2 and line 1.

Find the distance between line 3 and line 2.

Find the distance between line 3 and line 1.

Now to find out if it works, lets plug in our values into the Pythagorean Theorem:

The squared square roots cancel each other out and we are left with:

This proves that yes, the coordinates do indeed form a right triangle.