1. You should be familiar with the rectangular coordinate system and point plotting from an earlier algebra course. Let's just run through the basics. x axis y axis origin Quadrant I where both x and y are positive Quadrant II where x is negative and y is positive Quadrant III where both x and y are negative Quadrant IV where x is positive and y is negative

2. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 Let's plot the point (6,4) Let's plot the point (-3,-5) Let's plot the point (0,7) Let's plot the point (-6,0)

3. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 We now want to find the distance between two points. So the distance from (-6,4) to (1,4) is 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. (1,4)(-6,4) 7 units apart

4. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 What coordinate will be the same if the points are located vertically from each other? 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. (-6,-3)(-6,4) 7 units apart

5. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 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 add some lines and make a right triangle. This triangle measures 4 units by 3 units on the sides. If we find the hypotenuse, we'll have the distance from (0,0) to (4,3) Let's start by finding the distance from (0,0) to (4,3) ? 4 3 The Pythagorean Theorem will help us find the hypotenuse 5 So the distance between (0,0) and (4,3) is 5 units.

6. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 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. Let's add some lines and make a right triangle. Solving for c gives us: Let's start by finding the distance from (x 1,y 1 ) to (x2,y2)(x2,y2) ? x 2 - x 1 y 2 – y 1 Again the Pythagorean Theorem will help us find the hypotenuse (x 2,y 2 ) (x1,y1)(x1,y1) This is called the distance formula

7. Let's use it to find the distance between (3, -5) and (-1,4) (x1,y1)(x1,y1)(x2,y2)(x2,y2) 3 -5 4 CAUTION! You must do the parenthesis first then powers (square the numbers) and then add together BEFORE you can square root Don't forget the order of operations! means approximately equal to found with a calculator Plug these values in the distance formula

8. 2-7-6-5-4-3-21573 0468 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 If we have two points and we want to find the point halfway between them, it is called the midpoint. All this is, is an average of the x’s from the endpoints to get the x value of the midpoint and an average of the y’s from the endpoints to get the y value of the midpoint. This is easier to remember than memorizing a formula. Let's start by finding the midpoint between (0,0) and (4,3) First let’s look at the official formula for finding the midpoint. (4, 3) (0, 0)