Question

Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square.

Answer 1

Given points are (1,7),(4,2),(-1,-1),(-4,4)
Let the points are A,B,C,D.
Using the distance formula

AB = $\sqrt{(4-1{)}^2+(2-7{)}^2}$

AB = $\sqrt{9+25}=\sqrt{34}$

BC = $\sqrt{(-1-4{)}^2+(-1-2{)}^2}$

BC = $\sqrt{9+25}=\sqrt{34}$

CD = $\sqrt{(-4-(-1){)}^2+(4-(-1){)}^2}$
CD = $\sqrt{9+25}=\sqrt{34}$
DA = $\sqrt{(1-(-4){)}^2+(7-4){)}^2}$
DA = $\sqrt{9+25}=\sqrt{34}$

We know that the all sides of the square are equal.
Length of diagonal AC = $\sqrt{(-1-1{)}^2+(-1-7{)}^2}$

AC = $\sqrt{4+64}$ =$\sqrt{68}$

Length of diagonal BD = $\sqrt{(-4-4{)}^2+(4-2{)}^2}$
BD = $\sqrt{64+4}$ = $\sqrt{68}$

Length of diagonal BD= Length of diagonal AC
So,from this, we can say that these points are the vertices of a square.