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?

Question

asked Dec 24, 2020 91.4k views

2 Answers

The Ghost · Answer 1 · 2020-12-29T10:48:48+0000

Answer: A(2, 6) and C(2, -6)

Explanation:

The distance between any two points is given by :-

d=√((d-b)^2+(c-a)^2)

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

Then,

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

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

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

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

$AC=√((-6-6)^2+(2-2)^2)=√((-12)^2)=√(144)\\\\=\12$

$BD=√((0-0)^2+(-4-6)^2)=√((-10)^2)=√(100)\\\\=10$

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

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