Let's find the distance between two points. 8

Let's find the distance between two points. 8

Let's find the distance between two points. 8 7 units apart 7 (-6,4) 6 5 4 3 2 1 (1,4) So the distance from (-6,4) to (1,4) is 7. -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 If the points are located horizontally from each other, the y coordinates will be the same. You can look to see how far apart the x coordinates are. What coordinate will be the same if the points are located vertically from each other? (-6,4) 7 units apart 8 7

6 5 4 3 2 1 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 (-6,-3) So the distance from (-6,4) to (-6,-3) is 7. If the points are located vertically from each other, the x coordinates will be the same. You can look to see how far apart the y coordinates are. But what are we going to do if the points are not located either horizontally or vertically to find the distance between them? Let's start by finding the distance from (0,0) to (4,3) 8 7 6 5 4 3 2 1 2

a b c 5? 2 3 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 4 The Pythagorean Theorem will help us find the hypotenuse 2 2 2 4 3 c 2 16 9 c 2 c 5 So the distance between (0,0)

and (4,3) is 5 units. This triangle measures 4 units by 3 Let's add some lines and units on the sides. If we find the hypotenuse, we'll have the distance make a right triangle. from (0,0) to (4,3) Now let's generalize this method to come up with a formula so we don't have to make a graph and triangle every time. 8 7 6 2 5 4 3 2 11 Let's start by finding the distance from (x1,y1) to (x2,y ) (x2,y2) ? 2 2 a b c y2 y1 (x ,y1) x -x -7 -6 -5 -4 -3 -2 -1 0 1 2 32 4 5 16 7 8 -2

-3 2 -4 2 1 -5 -6 -7 Again the Pythagorean Theorem will help us find the hypotenuse Let's add some lines and make a right triangle. x x 2 y2 y1 c Solving for c gives us: c x2 x1 2 y2 y1 This is called the distance formula 2

2 2 Let's use it to find the distance between (3, -5) and (-1,4) Plug these values in the distance formula c 2 c 4 9 CAUTION! x-12 x31 2 (x1,y1) 2 y42 -5 y1 (x2,y2) 2 means approximately equal to 16 81 97 9.8 found with a calculator Don't forget the order of operations! You must do the brackets first then powers (square the numbers) and then add together BEFORE you can square root

Let's use it to find the distance between (6, -2) and (-2,2) Plug these values in the distance formula c x-22 x61 (x1,y1) 2 y22 -2 y1 (x2,y2) 2 Put Answer in simplest radical fo c 8 4 64 16 804 5 2 2 CAUTION! Don't forget the order of operations! You must do the brackets first then powers (square the numbers) and then add together BEFORE you can square root

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