Question

Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram​

Answer 1

Given :-

The vertices as P, Q, R, S such that

• P (2, - 2)

• Q (7, 3)

• R (11, - 1)

and

• S (6, - 6)

To Prove :-

• PQRS is a parallelogram.

Concept Used :-

We know,

In parallelogram, diagonals bisect each other.

So in order to prove that given vertices of P, Q, R, S taken in order forms a parallelogram, it is sufficient to show that midpoint of AC is equals to midpoint of BD.

$\green{\large\underline{\sf{Solution-}}}$

The vertices as P, Q, R, S such that

• P (2, - 2)

• Q (7, 3)

• R (11, - 1)

and

• S (6, - 6)

We know,

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Let us first find midpoint of PR.

• • Coordinates of P = ( 2, - 2)

• • Coordinates of Q = (11, - 1)

Using midpoint Formula,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR\: = \: \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

Here,

• x₁ = 2

• x₂ = 11

• y₁ = - 2

• y₂ = - 1

So,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{2 + 11}{2} , \dfrac{ - 2 - 1}{2} \bigg)$

$\rm :\longmapsto\:\boxed{\tt{ \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{13}{2} , \dfrac{ - 3}{2} \bigg)}} - - - (1)$

Now, To find Midpoint of QS

• • Coordinates of Q = (7, 3)

• • Coordinates of S = (6, - 6)

Using Midpoint Formula,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

Here,

• x₁ = 7

• x₂ = 6

• y₁ = 3

• y₂ = - 6

Thus,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{7+6}{2} , \dfrac{ - 6 + 3}{2} \bigg)$

$\rm :\longmapsto\: \boxed{\tt{ \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{13}{2} , \dfrac{ - 3}{2} \bigg)}} - - - (2)$

From equation (1) and (2), we get

$\bf\implies \:Midpoint \: of \: PR = Midpoint \: of \: QS$

$\bf\implies \:P,Q,R,S \: are \: the \: vertices \: of \: a \: parallelogram$

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More to Know :

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$