Question

If P(2,4), Q(0,3), R(3,6) and S(5,y) are the vertices of a parallelogram PQRS, then the value of y is
(a) 7 (b) 5
(c) -7 (d) -8

Subject: Mathematics
Chapter: Coordinate Geometry
Class: 10​

Answer 1

We know that the diagonals of the parallelogram bisect each other. So, O is the mid-point of PR and QS.

Coordinates of mid-point of PR = (² + ³, 4 + 0) =(², ²) - (² 5 10 2 2 5 22 2

Coordinates of mid-point of QS = 5 (0+5, 3+0) - (2, 3 + 9) 2 2

Now, these points coincides at the point O.

· (2, ³ + ") - ( ² , 0 ) 5 2² 3+ y 2 = (1/1, 5)

5

2

> 3+ y = 10

⇒y=7

Thus, the value of y is 7.

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Answer 2

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

Given that,

P(2,4), Q(0,3), R(3,6) and S(5,y) are the vertices of a parallelogram PQRS.

We know,

• In parallelogram, diagonals bisect each other.

So,

In order to find the value of y, we use the concept Midpoint of PR is equals go Midpoint of QS.

We know,

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:

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

➢ Let us first find midpoint of PR

Coordinates of P = (2, 4)

Coordinates of R = (3, 6)

Using midpoint Formula, we have

• x₁ = 2

• x₂ = 3

• y₁ = 4

• y₂ = 6

So,

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

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{2 + 3}{2} , \dfrac{4 + 6}{2} \bigg)$

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{5}{2} , \dfrac{10}{2} \bigg)$

$\rm :\longmapsto\: \sf \: Midpoint \: of \: PR \: = \: \bigg(\dfrac{5}{2} , 5 \bigg)$

➢ Now, Let's find the midpoint of QS

Coordinates of P = (0, 3)

Coordinates of R = (5, y)

Using midpoint Formula, we have

• x₁ = 0

• x₂ = 5

• y₁ = 3

• y₂ = y

So,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{0 + 5}{2} , \dfrac{3 + y}{2} \bigg)$

$\rm :\longmapsto\: \sf \: Midpoint \: of \: QS \: = \: \bigg(\dfrac{5}{2} , \dfrac{3 + y}{2} \bigg)$

Now, we have Midpoint of PR = Midpoint of QS

So,

$\rm :\longmapsto\: \sf \:\bigg(\dfrac{5}{2} , 5 \bigg) \: = \: \bigg(\dfrac{5}{2} , \dfrac{3 + y}{2} \bigg)$

So, on comparing, we get

$\rm :\longmapsto\:\dfrac{3 + y}{2} = 5$

$\rm :\longmapsto\:3 + y = 10$

$\rm :\longmapsto\: y = 10 - 3$

$\bf\implies \:y = 7$

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Learn More:

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:

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

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:

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

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:

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