Distance and Midpoint Lessons

Length

Distance

Line

Segment

quadratic formula

slope formula

Pythagorean theorem

none of these answers

$$d=\sqrt{\left(y_1-y_2\right)^2-\left(x_{2\ }-\ x_1\right)^2}\$$

$$d=\sqrt{\left(x_2-\ x_1\right)^2+\left(y_2-\ y_1\right)^2}$$

$$d=\sqrt{\left(y_1-y_2\right)^2+\ \left(x_{2\ }-\ x_1\right)^2}$$

$$d=\sqrt{\left(x\right)^2-\ \left(y\right)^2}$$

$$\sqrt{\left(-2-3\right)^2\ +\ \left(-1-1\right)^2}$$

$$\sqrt{\left(1-3\right)^2\ +\ \left(-1-2\right)^2}$$

$$\sqrt{\left(-2-3\right)^2\ -\ \left(-1-1\right)^2}$$

$$\sqrt{\left(-2-3\right)^2\ -\ \left(-1-1\right)^2}$$

$$d\ =\ \sqrt{\left(4-2\right)^2\ -\ \left(0-2\right)^2}$$

$$d\ =\ \sqrt{\left(4-0\right)^2\ -\ \left(-2-2\right)^2}$$

$$d\ =\ \sqrt{\left(-2\ +0\right)^2\ +\ \left(4+2\right)^2}$$

$$d\ =\ \sqrt{\left(-2-2\right)^2\ +\ \left(4-0\right)^2}$$

0

8

9

7

0

undefined

5

4

$$d\ =\ \sqrt{\left(5-3\right)^2\ +\ \left(0--2\right)^2}$$

$$d\ =\ \sqrt{\left(2-5\right)^2\ +\ \left(3\ -0\right)^2}$$

$$d\ =\ \sqrt{\left(5-3\right)^2\ -\ \left(0--2\right)^2}$$

$$d\ =\ \sqrt{\left(4-3\right)^2\ -\ \left(2-4\right)^2}$$

(2,5)

(4,4)

(4,3)

(1,0)

$$41$$  

$$\sqrt[]{38}$$  

$$\sqrt[]{41}$$  

$$\sqrt[]{25}$$  

Midpoint

Distance

Center

Line Segment

$$M=\left(3,3\right)$$  

$$M=\left(-3,-3\right)$$  

$$M=\left(-3,3\right)$$  

$$M=\left(3,-3\right)$$  

$$M=\left(-2,2\right)$$  

$$M=\left(2,-2\right)$$  

$$M=\left(-2,-2\right)$$  

$$M=\left(2,2\right)$$  

$$M=\left(-2,-\frac{1}{2}\right)$$  

$$M=\left(2,\frac{1}{2}\right)$$  

$$M=\left(-\frac{1}{2},-2\right)$$  

$$M=\left(\frac{1}{2},2\right)$$  

endpoint

midpoint

slope

distance

(3, 12)

(6, 4)

(1, 2)

(6, 8)

(-1, 3)

(2, -4)

(-6, 0)

( 6, 9)

(18, -9)

(5,-2)

(-5,2)

(6, 2)

(-1,4)