Question

Show that the points A(2, 1), B(5,2), C(6,4) and D(3, 3) are the angular
points of a parallelogram. Is this figure a rectangle?
step by step explaination​

Answer 1

Answer:

it cannot be rectangle,.........

Answer 2

Answer:

Parallelogram ABCD is not a rectangle.

Step-by-step explanation:

First let us give names to the points:

A(2,1)

B(5,2)

C(6,4)

D(3,3)

These are the vertices of the Parallelogram.

Solving using distance formula,

Distance Formula:

${\boxed{\boxed{\sf{\implies{\sqrt{(x2 - x1)^{2}}}}}}$

Now solving all by using this Formula:

A$C^{2}$

=> (6-2$)^{2}$ + (4-1$)^{2}$

=> (4$)^{2}$ +(3$)^{2}$

=> 16+9 = 25

B$C^{2}$ = (6-5$)^{2}$  + (4-2$)^{2}$

=> (1$)^{2}$ +(2$)^{2}$

=> 1+4 = 5

A$B^{2}$ = (5-2$)^{2}$ +(2-1$)^{2}$

=> (3$)^{2}$ +(1$)^{2}$

=> 9+1 = 10

D$C^{2}$

=> (6-3$)^{2}$ +(4-3$)^{2}$

=> (3$)^{2}$ +(1$)^{2}$  

=> 9+1=10

A$D^{2}$

=> (3-2$)^{2}$ +(3-1$)^{2}$

=> (1$)^{2}$ +(2$)^{2}$

=> 1+4= 5

Since BC = AD and DC = AB,

ABCD is a parallelogram.

A$B^{2}$   + B$C^{2}$  

=? 10+5 = 15

A$B^{2}$  + B$C^{2}$  ≠ A$C^{2}$

∴△ABC is not right angled.

Therefore, Parallelogram ABCD is not a rectangle.