Question

Is my rectangle right or wrong

Answer 1

Heya !!!

Here is your answer.

Let us first calculate the distance.

Using distance formula,
$PQ = \sqrt{{(x2 - x1)}^{2} + {(y2 - y1) }^{2} }$
So,
$AB= \sqrt{{(6 - 0 )}^{2} + {(7 + 1) }^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} = 10 \: units$
$BC = \sqrt{{( - 2 - 6)}^{2} + {(3 - 7) }^{2} } \\ = \sqrt{ { (- 8)}^{2} + { (- 4)}^{2} } \\ = \sqrt{64 + 16} = \sqrt{80} \: units$
$CD = \sqrt{{(8 + 2)}^{2} + {(3 - 3) }^{2} } \\ = \sqrt{ {(10)}^{2} } \\ = \sqrt{100} = 10 \: units$
$DA = \sqrt{{(0 - 8)}^{2} + {( - 1 - 3) }^{2} } \\ = \sqrt{ { (- 8)}^{2} + { (- 4)}^{2} } \\ = \sqrt{64 + 16} = \sqrt{80} \: units$
Now, Diagonals...
$AC = \sqrt{{( - 2 - 0)}^{2} + {(3 + 1) }^{2} } \\ = \sqrt{ {( - 2)}^{2} + {(4)}^{2} } \\ = \sqrt{4 + 16} = \sqrt{20} \: units$
$BD = \sqrt{{(8 - 6)}^{2} + {(3 -7) }^{2} }\\ = \sqrt{ {( 2)}^{2} + {( - 4)}^{2} } \\ = \sqrt{4 + 16} = \sqrt{20} \: units$

Now,
Opposite Sides...
AB = CD = 10 units.
BC = DA = √80 units.

And, Diagonals...
AC = BD = √20 units.

Therefore, it is a rectangle.

Hope It Helps