Answer:
(7, 3), (–1, –1) and (2, 8) are at a distance of units from (2, 3).
Step-by-step explanation:
We know that the formula of distance is given by:
Distance = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
So substituting the given points points to check which points are at a distance of 5 units from [tex](2,3)[/tex].
1. (2, 3) and (7, 3):
[tex]\sqrt{(7-2)^2+(3-3)^2} =\sqrt{25} = 5[/tex]
2. (2, 3) and (–1, –1):
[tex]\sqrt{(-1-2)^2+(-1-3)^2} =\sqrt{25} = 5[/tex]
3. (2, 3) and (0, 3):
[tex]\sqrt{(0-2)^2+(3-3)^2} =\sqrt{4} = 2[/tex]
4. (2, 3) and (2, 8):
[tex]\sqrt{(2-2)^2+(8-3)^2} =\sqrt{25} = 5[/tex]
5. (2, 3) and (-7, 6):
[tex]\sqrt{(-7-2)^2+(6-3)^2} =\sqrt{90} =9.5[/tex]
