Question

show that the quadrilateral whose vertices are (3,8) ,(7,6) ,(3,-2) , and (-1,0) is a rectangle​

Answer 1

[ Note : refer the figure in the attachment ]

$\boxed{\textbf{\large{To prove}}}$

The quadrilateral whose vertices are (3,8) ,(7,6) ,(3,-2) , and (-1,0) is a rectangle

$\boxed{\textbf{\large{ Proof }}}$

◾let us consider the quadrilateral is ⬛ ABCD, vertices are of quadrilateral are,

A(3,8) = (x1, y1)

B(7,6) = (x2 , y2)

C(3,-2)= (x3 , y3 )

D(-1,0)= ( x4 , y4)

◾So, first of all, we have to find the distance of sideAB and sideCD

◾we know, the distance formula

Distance = √[(y2 - y1)^2 + (x2 - x1 )^2]

◾let us find the distance of AB

side AB = √[ (6 - 8)^2 + ( 7 - 3)^2 ]

= √[ ( 2)^2 + (4)^2 ]

= √ [ 4 + 16 ]

= √ 20

= 2 √ 5

Now, find DC

side DC=√[( -2 - (0))^2 + ( 3 - (-1))^2]

=√ [(-2)^2 + (4)^2 ]

=√ [ 4 + 16 ]

= √ 20

= 2√5

◾Now find, the lengths of sideAD and sideBC

side AD =√[(0-8)^2 + (-1-3)^2]

=√[ (-8)^2 + (-4)^2 ]

= √ [ 64 + 16 ]

=√80

= 4√5

Now, find BC

side BC =√[ ( -2-6)^2 + (3-7)^2 ]

=√[ (-8)^2 + (-4)^2 ]

=√[ 64 + 16 ]

=√ 80

= 4√5

◾From above, we can conclude

that

side AB = side DC And

side AD = side BC

◾from above, it satisfy the properties of parallelogram, for the rectangle we also have to show the diagonals of a quadrilateral are equal in magnitude

◾Therefor the diagonals of quadrilateral are diagonalAC and diagonal BD

◾let us find Diagonals,.

Diagonal AC

=√ [ (( -2)-(8))^2 + (3 - 3 )^2 ]

=√ [ ( -10)^2 ]

=√ 100

= 10

◾Now, find diagonal BD

Diagonal BD

=√ [ ( 0 - 6)^2 + ( -1 - 7 )^2 ]

=√ [ (-6)^2 + ( -8 )^2 ]

=√ [ 36 + 64 ]

=√ [ 100 ]

= 10

So,

◾the Diagonal BD = Diagonal AC

And also the opposite sides of quadrilateral are equal ( side AB = side DC, Side AD = Side BC)

◾from above, As we consider the properties of rectangle , given ⬛ABCD satisfied the properties of rectangle

Therefor ⬛ABCD is a rectangle

_________________________________-

Answer 2

Step-by-step explanation:

let us consider the quadrilateral is ⬛ ABCD, vertices are of quadrilateral are,

A(3,8) = (x1, y1)

B(7,6) = (x2 , y2)

C(3,-2)= (x3 , y3 )

D(-1,0)= ( x4 , y4)

◾So, first of all, we have to find the distance of sideAB and sideCD

◾we know, the distance formula

Distance = √[(y2 - y1)^2 + (x2 - x1 )^2]

◾let us find the distance of AB

side AB = √[ (6 - 8)^2 + ( 7 - 3)^2 ]

= √[ ( 2)^2 + (4)^2 ]

= √ [ 4 + 16 ]

= √ 20

= 2 √ 5

Now, find DC

side DC=√[( -2 - (0))^2 + ( 3 - (-1))^2]

=√ [(-2)^2 + (4)^2 ]

=√ [ 4 + 16 ]

= √ 20

= 2√5

◾Now find, the lengths of sideAD and sideBC

side AD =√[(0-8)^2 + (-1-3)^2]

=√[ (-8)^2 + (-4)^2 ]

= √ [ 64 + 16 ]

=√80

= 4√5

Now, find BC

side BC =√[ ( -2-6)^2 + (3-7)^2 ]

=√[ (-8)^2 + (-4)^2 ]

=√[ 64 + 16 ]

=√ 80

= 4√5

◾From above, we can conclude

that

side AB = side DC And

side AD = side BC

◾from above, it satisfy the properties of parallelogram, for the rectangle we also have to show the diagonals of a quadrilateral are equal in magnitude

◾Therefor the diagonals of quadrilateral are diagonalAC and diagonal BD

◾let us find Diagonals,.

Diagonal AC

=√ [ (( -2)-(8))^2 + (3 - 3 )^2 ]

=√ [ ( -10)^2 ]

=√ 100

= 10

◾Now, find diagonal BD

Diagonal BD

=√ [ ( 0 - 6)^2 + ( -1 - 7 )^2 ]

=√ [ (-6)^2 + ( -8 )^2 ]

=√ [ 36 + 64 ]

=√ [ 100 ]

= 10

So,

◾the Diagonal BD = Diagonal AC

And also the opposite sides of quadrilateral are equal ( side AB = side DC, Side AD = Side BC)

◾from above, As we consider the properties of rectangle , given ⬛ABCD satisfied the properties of rectangle

Therefor ⬛ABCD is a rectangle