Question

date: per: homework 3: distance & midpoint this is a 2 - page document! directions: find the distance between each pair of points. 1. (-4, 6) and (3, -7) 2. (-6, -5) and (2, 0) 3. (-1, 4) and (1, -1) 4. (0, -8) and (3, 2)

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Solve for problem 1

Let $(x_1,y_1)=(-4,6)$ and $(x_2,y_2)=(3,-7)$. Then $d_1=\sqrt{(3 - (-4))^2+((-7)-6)^2}=\sqrt{(3 + 4)^2+(-13)^2}=\sqrt{49 + 169}=\sqrt{218}$.

Step3: Solve for problem 2

Let $(x_1,y_1)=(-6,-5)$ and $(x_2,y_2)=(2,0)$. Then $d_2=\sqrt{(2-(-6))^2+(0 - (-5))^2}=\sqrt{(2 + 6)^2+5^2}=\sqrt{64 + 25}=\sqrt{89}$.

Step4: Solve for problem 3

Let $(x_1,y_1)=(-1,4)$ and $(x_2,y_2)=(1,-1)$. Then $d_3=\sqrt{(1-(-1))^2+((-1)-4)^2}=\sqrt{(1 + 1)^2+(-5)^2}=\sqrt{4 + 25}=\sqrt{29}$.

Step5: Solve for problem 4

Let $(x_1,y_1)=(0,-8)$ and $(x_2,y_2)=(3,2)$. Then $d_4=\sqrt{(3 - 0)^2+(2-(-8))^2}=\sqrt{3^2+(2 + 8)^2}=\sqrt{9+100}=\sqrt{109}$.

Answer:

1. $\sqrt{218}$
2. $\sqrt{89}$
3. $\sqrt{29}$
4. $\sqrt{109}$