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

use midpoint formula

AC 1-1/2,7-1/2

0/2,6/2

0,3

BD 4-4/2,2+4/2

0/2,6/2

0,3

As AC and BD are same ,hence proved...

Hope it helps u...

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

A(1, 7)

B (4, 2)

C (-1, -1)

D (- 4, 4)

Using Distance Formula :-

AB = √(x2 - x1)² + (y2 - y1)²

AB = √(4 - 1)² + (2 - 7)²

AB = √3² + (-5)²

AB = √9 + 25

AB = √34 units

Now, again

Using Distance Formula :-

BC= √(x2 - x1)² + (y2 - y1)²

BC = √(- 1 - 4)² + ( - 2 - 1)²

BC = √(-5)² + (-3)²

BC = √25 + 9

BC = √34 units

Now, Again

Using Distance Formula :-

CD = √(x2 - x1)² + (y2 - y1)²

CD = √(- 4 + 1)² + (4 + 1)²

CD = √(-3)² + (5)²

CD = √9 + 25

CD = √34 units

Also,

Using Distance Formula :-

DA = √(x2 - x1)² + (y2 - y1)²

DA = √(- 4 - 1)² + (4 - 7)²

DA = √(-5)² + 3²

DA = √25 + 9

DA = √34 units

Now, Diagonal

Using Distance Formula :-

BD = √(x2 - x1)² + (y2 - y1)

BD = √(- 4 - 4)² + (4 - 2)²

BD = √(-8)² + 2²

BD = √64 + 4

BD = √68 units

Now, Diagonal

Using Distance Formula :-

AC = √(x2 - x1)² + (y2 - y1)

AC = √(- 1 - 1)² + (-1 - 7)²

AC = √(-2)² + (-8)²

AC = √4 + 64

AC = √68 units

Therefore, we get

AB = BC = CD = DA = √34

Also,

BD = AC = √68

Hence,

Proved