Answer: A(2, 6) and C(2, -6)
Step-by-step explanation:
The distance between any two points is given by :-
[tex]d=\sqrt{(d-b)^2+(c-a)^2}[/tex]
Given vertices : A(2, 6), B(6, 0), C(2, -6) and D(-4, 0)
Then,
[tex]AB=\sqrt{(0-6)^2+(6-2)^2}=\sqrt{(-6)^2+(4)^2}=\sqrt{36+16}\\\\=\sqrt{52}\approx7.211[/tex]
[tex]BC=\sqrt{(-6-0)^2+(2-6)^2}=\sqrt{(-6)^2+(-4)^2}=\sqrt{36+16}\\\\=\sqrt{52}\approx7.211[/tex]
[tex]CD=\sqrt{(0-(-6))^2+(-4-2)^2}=\sqrt{(6)^2+(-6)^2}=\sqrt{36+36}\\\\=\sqrt{72}\approx8.485[/tex]
[tex]AD=\sqrt{(0-2)^2+(-4-6)^2}=\sqrt{(-2)^2+(-10)^2}=\sqrt{4+100}\\\\=\sqrt{104}\approx10.198[/tex]
[tex]AC=\sqrt{(-6-6)^2+(2-2)^2}=\sqrt{(-12)^2}=\sqrt{144}\\\\=\12[/tex]
[tex]BD=\sqrt{(0-0)^2+(-4-6)^2}=\sqrt{(-10)^2}=\sqrt{100}\\\\=10[/tex]
Since, points A and C has the largest distance between them.
Therefore, the points A and C are the farthest points.
