Question

Show that points (-2,-1), (4,0), (3,3) and (-3,2) are the vertices of a parallelogram.​

Answer 1

The points (-2,-1), (4,0), (3,3) and (-3,2) are the vertices of a parallelogram, "proved".​

Step-by-step explanation:

Let  A(- 2, - 1), B(4, 0), C(3, 3) and D(- 3, 2) are the vertices of a parallelogram.

Here, A(x_1 = - 2, y_1 = - 1), B(x_2 = 4, y_2 = 0), C(x_3 = 3, y_3 = 3) and

D( x_4 = - 3, y_4 =2)

∴ $AB^{2}$ = $(4 + 2)^{2} + (0 + 1)^{2}$ = 36 + 1 = 37

$BC^{2}$ = $(3 - 4)^{2} + (3 - 0)^{2}$ = 1 + 9 = 10

$CD^{2}$ = $(- 3 - 3)^{2} + (2 - 3)^{2}$ = 36 + 1 = 37

and $DA^{2}$ = $(- 3 + 2)^{2} + (2 + 1)^{2}$ = 1 + 9 = 10

Also,

$AC^{2}$ = $(3 + 2)^{2} + (3 + 1)^{2}$ = 25 + 9 = 34

$BD^{2}$ = $( - 3 - 4)^{2} + (2 - 0)^{2}$ = 49 + 4 = 53

∴ $AB^{2}$ = $CD^{2}$ = 37 and

$BC^{2}$ =  $DA^{2}$

Opopsite sides are equal .

and diagonal are not equal, it is a parallelogram.

Hence, it is proved.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

P(-2 , -1)

Q(4 , 0)

R(3 , 3)

S(-3 , 2)

$\textbf{\underline{These\;are\;the\;vertices\;of\; Quadrilateral\;PQRS.}}$

Now,

SLOPE PQ

${\boxed{\sf\:{\dfrac{0-(-1)}{4-(2)}}}}$

${\boxed{\sf\:{\dfrac{0+1}{4+2}}}}$

${\boxed{\sf\:{\dfrac{1}{6}}}}$

SLOPE SR

${\boxed{\sf\:{\dfrac{3-2}{3-(-3)}}}}$

${\boxed{\sf\:{\dfrac{1}{3+3}}}}$

${\boxed{\sf\:{\dfrac{1}{6}}}}$

SLOPE QR

${\boxed{\sf\:{\dfrac{3-0}{3-4}}}}$

${\boxed{\sf\:{\dfrac{3}{-1}}}}$

= -3

SLOPE PS

${\boxed{\sf\:{\dfrac{2-(-1)}{-3-(-2)}}}}$

${\boxed{\sf\:{\dfrac{2+1}{-3+2}}}}$

${\boxed{\sf\:{\dfrac{3}{-1}}}}$

= -3

Therefore,

PQ = SR

PQ || SR

Also,

QR = PS

QR || PS

Hence,

$\textbf{\underline{PQRS\;is\;a\;parallelogram}}$