Question

1. 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 $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here $(x_1,y_1)=(2,2)$ and $d = 10$.

Step2: Check option a

For point $(7,3)$:
$d=\sqrt{(7 - 2)^2+(3 - 2)^2}=\sqrt{25 + 1}=\sqrt{26}
eq10$.

Step3: Check option b

For point $(-6,-4)$:
$d=\sqrt{(-6 - 2)^2+(-4 - 2)^2}=\sqrt{(-8)^2+(-6)^2}=\sqrt{64 + 36}=\sqrt{100}=10$.

Step4: Check option c

For point $(5,0)$:
$d=\sqrt{(5 - 2)^2+(0 - 2)^2}=\sqrt{9+4}=\sqrt{13}
eq10$.

Step5: Check option d

For point $(-3,-1)$:
$d=\sqrt{(-3 - 2)^2+(-1 - 2)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25 + 9}=\sqrt{34}
eq10$.

Answer:

b. $(-6,-4)$