Question

for each of the following, find the distance between the points. 18. (3, 8), (9, 10) 19 (6, 4), (-5, -1) 20 (-5, 6), (8, -4) 21. 22. 23. for each of the following, find the perimeter of the shape. 24. 25. 26 point a is located at (2,2) and the length of the segment is 10 units. which of the following could be the other endpoint? a (7, 3) b. (-6, -4) c (5, 0) d (-3, -1)

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 problem 18

For points $(3,8)$ and $(9,10)$, let $(x_1,y_1)=(3,8)$ and $(x_2,y_2)=(9,10)$. Then $d=\sqrt{(9 - 3)^2+(10 - 8)^2}=\sqrt{6^2+2^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}$.

Step3: Solve problem 19

For points $(6,4)$ and $(-5,-1)$, let $(x_1,y_1)=(6,4)$ and $(x_2,y_2)=(-5,-1)$. Then $d=\sqrt{(-5 - 6)^2+(-1 - 4)^2}=\sqrt{(-11)^2+(-5)^2}=\sqrt{121+25}=\sqrt{146}$.

Step4: Solve problem 20

For points $(-5,6)$ and $(8,-4)$, let $(x_1,y_1)=(-5,6)$ and $(x_2,y_2)=(8,-4)$. Then $d=\sqrt{(8+5)^2+(-4 - 6)^2}=\sqrt{13^2+(-10)^2}=\sqrt{169 + 100}=\sqrt{269}$.

Step5: Solve problem 26

For point $A(2,2)$ and an endpoint $(x,y)$, using the distance formula $d = 10=\sqrt{(x - 2)^2+(y - 2)^2}$.
For option a: $(x,y)=(7,3)$, $d=\sqrt{(7 - 2)^2+(3 - 2)^2}=\sqrt{25 + 1}=\sqrt{26}
eq10$.
For option b: $(x,y)=(-6,-4)$, $d=\sqrt{(-6 - 2)^2+(-4 - 2)^2}=\sqrt{64+36}=\sqrt{100}=10$.
For option c: $(x,y)=(5,0)$, $d=\sqrt{(5 - 2)^2+(0 - 2)^2}=\sqrt{9 + 4}=\sqrt{13}
eq10$.
For option d: $(x,y)=(-3,-1)$, $d=\sqrt{(-3 - 2)^2+(-1 - 2)^2}=\sqrt{25 + 9}=\sqrt{34}
eq10$.

Answer:

1. $2\sqrt{10}$
2. $\sqrt{146}$
3. $\sqrt{269}$
4. B. $(-6,-4)$