Question

A quadrilateral has vertices (2 + √2, −1), (8 + √2, 3), (6 + √2, 6), and (√2, 2). Prove that the quadrilateral is a

rectangle.

Answer 1

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!

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:

`r = √((x_1-x^2)^2+(y_1-y^2)^2)`

distance AB

`\overline{AB} = \sqrt{(2+√(2)-8-√(2))^2+(-1-3)^2}`

`\overline{AB} = √((2-8)^2+(-1-3)^2)`

`\overline{AB} = √(36+16) = √(52)`

distance BC

`\overline{BC} = \sqrt{(8+√(2)-(6+√(2)))^2+(3-6)^2}`

`\overline{BC} = √(13)`

since these two distances are different, the next two SHOULD be equal to AB and BC

distance CD

`\overline{CD} = \sqrt{(6+√(2)-√(2))^2+(6-2)^2}`

`\overline{CD} = √(52) = \overline{AB}`

distance DA

`\overline{DA} = \sqrt{(√(2)-(2+√(2)))^2+(-2-1)^2}`

`\overline{DA} = √(13) = \overline{BC}`

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

`m= (y_2 - y_1)/(x_2-x_1)`

checking: slope of AB = slope of CD:

`AB_m= CD_m`

`(3-(-1))/(8+√(2)-(2+√(2)))=(2 - 6)/(√(2)-(6+√(2)))`

`(3+1)/(8-2)=(2-6)/(-6)`

`(4)/(6)=(-4)/(-6)`

`(4)/(6)=(4)/(6)`

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

`BC_m= DA_m`

`(6-3)/((6+√(2))-(8+√(2)))=(-1-2)/((2+√(2))-(√(2)))`

`(-3)/(2)=(-3)/(2)`

hence the lines BC and DA are parallel and equal as well!