Question

Show that the points A(5, 6), B (1, 5), C(2, 1) and D(6, 2) are the vertices of a square.

Answer 1

Answer:

HOLA MATE YOUR ANSWER 》》》》》

Answer 2

To prove :

Points A(5,6),B(1,5),C(2,1) & D(6,2) are the vertices of the square.

Formula used :

Distance formula,

$\sqrt{ ({x2 - x1}^{2}) + {(y2 - y1)}^{2} }$

Here (x1, y1) and (x2, y2) are the coordinated of a line segment.

Proof :

For side AB,

A(5,6) & B(1,5)

$AB \: = \sqrt{ {(1 - 5)}^{2} + {(5 - 6)}^{2} }$

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

$AB \: = \sqrt{16 + 1} = \sqrt{17}$

And for side CD,

C(2,1) & D(6,2)

$CD \: = \sqrt{ {(6 - 2}^{2} + {(2 - 1)}^{2} }$

$CD \: = \sqrt{ {( 4)}^{2} + {(1)}^{2} }$

$CD \: = \sqrt{16 + 1} = \sqrt{17}$

As AB =CD

similarly the value of all the sides are equal.

$AB =BC=CD=AD= \sqrt{17}$

Hence, proved that it is a square.