Question

Three vertices of a rhombus PQRS

are P(2, -1) , Q( 3, 4) , R( -2, 3) find the

coordinates of the fourth vertex S​

Answer 1

Given :-

A rhombus PQRS whose three vertices are

• P (2, -1)

• Q (3, 4)

• R (- 2, 3)

To Find :-

• The coordinates of the vertex S.

Concept Used :-

• Since, Rhombus is a parallelogram and we know, in parallelogram diagonals bisect each other.

Solution :-

Given three vertices of the Rhombus PQRS

• Coordinates of P be (2, - 1)

• Coordinates of Q be (3, 4)

• Coordinates of R be (- 2, 3).

Let

• Coordinates of S be (x, y).

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, - 1)

• Coordinates of R = (- 2, 3)

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 = - 2

• y₁ = - 1

• y₂ = 3

So,

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

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{0}{2} , \: \dfrac{2}{2} \bigg)$

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(0 , \:1 \bigg)$

Now,

Let us find midpoint of QS.

• Coordinates of Q = ( 3, 4)

• Coordinates of S = (x, y)

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 = 3

• x₂ = x

• y₁ = 4

• y₂ = y

So,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{3 + x}{2} , \dfrac{4 + y}{2} \bigg)$

Now,

We know that,

In rhombus, diagonals bisect each other

So, Midpoint of PR = Midpoint of QS

$\rm :\longmapsto\: \sf \: (0,1)= \: \bigg(\dfrac{3 + x}{2} , \dfrac{4 + y}{2} \bigg)$

On comparing, we get

$\rm :\longmapsto\:0 = \dfrac{3 + x}{2} \: \: and \: 1 = \dfrac{4 + y}{2}$

$\rm :\longmapsto\:3 + x = 0 \: \: \: and \: \: \: 4 + y = 2$

$\rm :\implies\:x = - 3 \: \: \: and \: \: \: y = - \: 2$

$\underbrace{ \boxed{ \bf \: Hence, \: Coordinates \: of \: S \: be \: ( - 3, - 2)}}$

Additional Information :-

Distance Formula :-

Distance formula is used to find the distance between two given Points

${\underline{\boxed{\rm{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

Section Formula :-

Section Formula is used to find the co ordinates of the point A(x, y) which divides the line segment joining the points (B) and (C) internally in the ratio m : n,

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} ,\dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$