Question

Show that the points (-3, 1), (-6, -7), (3, -9). (6.-1) taken in order form a parallelogram​

Answer 1

$\fontsize{18}{10}{\textup{\textbf{The proof is done below.}}}$

Step-by-step explanation:

Let the given points be represented by  A(-3, 1), B(-6, -7), C(3, -9) and D (6.-1).

We know that if a pair of opposite sides of a quadrilateral are parallel and congruent, then it is a PARALLELOGRAM.

The lengths of the opposite sides AB and CD are calculated, using distance formula, as follows :

$AB=\sqrt{(-6+3)^2+(-7-1)^2}=\sqrt{9+64}=\sqrt{73},\\\\CD=\sqrt{(6-3)^2+(-1+9)^2}=\sqrt{9+64}=\sqrt{73}.$

And, the slopes of the sides AB and CD are calculated as follows :

$\textup{slope of AB}=\dfrac{-1-7}{-6+3}=\dfrac{-8}{-3}=\dfrac{8}{3},\\\\\\\textup{slope of CD}=\dfrac{-1+9}{6-3}=\dfrac{8}{3}.$

Since the lengths and slopes of a pair of opposite sides AB and CD are equal, so ABCD is a parallelogram.

Hence showed.  

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Question : What are the properties of parallelogram?

Link: https://brainly.in/question/3062992

Answer 2

Step-by-step explanation:

Let the given points be represented by A(-3, 1), B(-6, -7), C(3, -9) and D (6.-1).

We know that if a pair of opposite sides of a quadrilateral are parallel and congruent, then it is a PARALLELOGRAM.

The lengths of the opposite sides AB and CD are calculated, using distance formula, as follows :

\begin{gathered}AB=\sqrt{(-6+3)^2+(-7-1)^2}=\sqrt{9+64}=\sqrt{73},\\\\CD=\sqrt{(6-3)^2+(-1+9)^2}=\sqrt{9+64}=\sqrt{73}.\end{gathered}

AB=

(−6+3)

2

+(−7−1)

2

=

9+64

=

73

,

CD=

(6−3)

2

+(−1+9)

2

=

9+64

=

73

.

And, the slopes of the sides AB and CD are calculated as follows :

\begin{gathered}\textup{slope of AB}=\dfrac{-1-7}{-6+3}=\dfrac{-8}{-3}=\dfrac{8}{3},\\\\\\\textup{slope of CD}=\dfrac{-1+9}{6-3}=\dfrac{8}{3}.\end{gathered}

slope of AB=

−6+3

−1−7

=

−3

−8

=

3

8

,

slope of CD=

6−3

−1+9

=

3

8

.

Since the lengths and slopes of a pair of opposite sides AB and CD are equal, so ABCD is a parallelogram.

Hence,proved