Question

Show that the points (-2, -1) (1, 0) (4, 3) and (1, 2) taken in order an vertices of parallelogram?

Answer 1

Answer:

Yes, the given points form a parallelogram.

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Step-by-step explanation:

For any 4 points to form a parallelogram, the lengths of the opposite sides need to be equal.

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Let us name the points (-2, -1), (1, 0), (4, 3) and (1, 2) A, B, C and D respectively. Where AB is opposite to DC, and BC is opposite to AD. [Refer to the diagram]

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In order to check if they're equal, we'll need to find out their lengths, and we'll find out the lengths with the help of the Distance formula.

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For any two points (x₁, y₁) and (x₂, y₂), the distance between the two points is given by;

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$\sf Distance \ Formula = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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For AB;

A = (-2, -1)

B = (1, 0)

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$\sf \dashrightarrow \ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow \ AB = \sqrt{(1 - (-2))^2 + (0 - (-1))^2}$

$\sf \dashrightarrow \ AB = \sqrt{(1 + 2)^2 + (0 + 1)^2}$

$\sf \dashrightarrow \ AB = \sqrt{(3)^2 + (1)^2}$

$\sf \dashrightarrow \ AB = \sqrt{9 + 1}$

$\sf \dashrightarrow \ AB = \sqrt{10} \ units.$

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For BC;

B = (1, 0)

C = (4, 3)

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$\sf \dashrightarrow \ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow \ BC = \sqrt{(4 - 1)^2 + (3 - 0)^2}$

$\sf \dashrightarrow \ BC = \sqrt{(3)^2 + (3)^2}$

$\sf \dashrightarrow \ BC = \sqrt{9 + 9}$

$\sf \dashrightarrow \ BC = \sqrt{18} \ units.$

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For CD;

C = (4, 3)

D = (1, 2)

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$\sf \dashrightarrow \ CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow \ CD = \sqrt{(1 - 4)^2 + (2 - 3)^2}$

$\sf \dashrightarrow \ CD = \sqrt{(-3)^2 + (-1)^2}$

$\sf \dashrightarrow \ CD = \sqrt{9 + 1}$

$\sf \dashrightarrow \ CD = \sqrt{10} \ units.$

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For AD;

A = (-2, -1)

D = (1, 2)

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$\sf \dashrightarrow \ AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow \ AD = \sqrt{(1 - (-2))^2 + (2 - (-1))^2}$

$\sf \dashrightarrow \ AD = \sqrt{(1 + 2)^2 + (2 + 1)^2}$

$\sf \dashrightarrow \ AD = \sqrt{(3)^2 + (3)^2}$

$\sf \dashrightarrow \ AD = \sqrt{9 + 9}$

$\sf \dashrightarrow \ AD = \sqrt{18} \ units.$

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On comparing the values of AB, BC, CD and DA, we can see that AB = CD and BC = DA.

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Therefore ABCD forms a parallelogram.

Answer 2

Answer:

We have the points

P(−2,−1),Q(1,0),R(4,3) and S(1,2)

We know the property the of parallelogram that diagonals of parallelogram bisect each other.

Let us find out mid-point of line joining P and R and line joining Q and S  

(i) Mid-point M of diagonal PR

M(  

2

−2+4

​

,  

2

−1+3

​

)

⇒M(1,1)

(ii) Mid-point M  

′

 of diagonal QS

M  

′

(  

2

1+1

​

,  

2

0+2

​

)

⇒M  

′

(1,1)

From (i) & (ii)

Mid-points M & M  

′

 are identical

⇒ Diagonals of the figure PQRS bisect each other and this property is enough to  prove that it is a parallelogram.

Although we can also check by distance formula i.e.  d=  

(a−c)  

2

+(b−d)  

2

 

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PQ=RS

SP=QR

Step-by-step explanation: