Question

A quadrilateral is graphed on a coordinate plane. If the vertices are A(2, 6), B(6, 0), C(2, –6) and D(–4, 0), which two points are farthest apart?

Answer 1

Check the picture below.

Answer 2

Answer:  A(2, 6) and C(2, -6)

Step-by-step explanation:

The distance between any two points is given by :-

$d=\sqrt{(d-b)^2+(c-a)^2}$

Given vertices : A(2, 6), B(6, 0), C(2, -6) and D(-4, 0)

Then,

$AB=\sqrt{(0-6)^2+(6-2)^2}=\sqrt{(-6)^2+(4)^2}=\sqrt{36+16}\\\\=\sqrt{52}\approx7.211$

$BC=\sqrt{(-6-0)^2+(2-6)^2}=\sqrt{(-6)^2+(-4)^2}=\sqrt{36+16}\\\\=\sqrt{52}\approx7.211$

$CD=\sqrt{(0-(-6))^2+(-4-2)^2}=\sqrt{(6)^2+(-6)^2}=\sqrt{36+36}\\\\=\sqrt{72}\approx8.485$

$AD=\sqrt{(0-2)^2+(-4-6)^2}=\sqrt{(-2)^2+(-10)^2}=\sqrt{4+100}\\\\=\sqrt{104}\approx10.198$

$AC=\sqrt{(-6-6)^2+(2-2)^2}=\sqrt{(-12)^2}=\sqrt{144}\\\\=\12$

$BD=\sqrt{(0-0)^2+(-4-6)^2}=\sqrt{(-10)^2}=\sqrt{100}\\\\=10$

Since, points A and C has the largest distance between them.

Therefore, the points A and C are the farthest points.