Respuesta :
Answer:
As the opposite lines have the same slope and are equal in lengths. It is proved that the quadrilateral with given points is indeed a rectangle!
Step-by-step explanation:
The condition for a quadrilateral to be a rectangle is that
- the opposite lines must be parallel.
- the opposite distances must be equal
Given the points A(2 + √2, −1), B(8 + √2, 3), C(6 + √2, 6), and D(√2, 2)
We don't know in what order they are given, so our first step should be to calculate the distances using the distance formula:
[tex]r = \sqrt{(x_1-x^2)^2+(y_1-y^2)^2}[/tex]
distance AB
[tex]\overline{AB} = \sqrt{(2+\sqrt{2}-8-\sqrt{2})^2+(-1-3)^2}[/tex]
[tex]\overline{AB} = \sqrt{(2-8)^2+(-1-3)^2}[/tex]
[tex]\overline{AB} = \sqrt{36+16} = \sqrt{52}[/tex]
distance BC
[tex]\overline{BC} = \sqrt{(8+\sqrt{2}-(6+\sqrt{2}))^2+(3-6)^2}[/tex]
[tex]\overline{BC} = \sqrt{13}[/tex]
since these two distances are different, the next two SHOULD be equal to AB and BC
distance CD
[tex]\overline{CD} = \sqrt{(6+\sqrt{2}-\sqrt{2})^2+(6-2)^2}[/tex]
[tex]\overline{CD} = \sqrt{52} = \overline{AB}[/tex]
distance DA
[tex]\overline{DA} = \sqrt{(\sqrt{2}-(2+\sqrt{2}))^2+(-2-1)^2}[/tex]
[tex]\overline{DA} = \sqrt{13} = \overline{BC}[/tex]
Now we have an idea that since AB = CD and BC = DA, then if this shape is a rectangle then these opposite lines should also be parallel to each other!
Hence we need to see if:
slope of AB = slope of CD
and
slope of BC = slope of DA
to find the slope between two points, we'll use
[tex]m= \dfrac{y_2 - y_1}{x_2-x_1}[/tex]
checking: slope of AB = slope of CD:
[tex]AB_m= CD_m[/tex]
[tex]\dfrac{3-(-1)}{8+\sqrt{2}-(2+\sqrt{2})}=\dfrac{2 - 6}{\sqrt{2}-(6+\sqrt{2})}[/tex]
[tex]\dfrac{3+1}{8-2}=\dfrac{2-6}{-6}[/tex]
[tex]\dfrac{4}{6}=\dfrac{-4}{-6}[/tex]
[tex]\dfrac{4}{6}=\dfrac{4}{6}[/tex]
hence the lines AB and CD are parallel and equal!
we can do the same for BC and DA
slope of BC = slope of DA
[tex]BC_m= DA_m[/tex]
[tex]\dfrac{6-3}{(6+\sqrt{2})-(8+\sqrt{2})}=\dfrac{-1-2}{(2+\sqrt{2})-(\sqrt{2})}[/tex]
[tex]\dfrac{-3}{2}=\dfrac{-3}{2}[/tex]
hence the lines BC and DA are parallel and equal as well!
