Question

A (3, 3), B (6, y) c (x7) and D (5, 6) are the vertices of a
parallelogram ABCD. Find x and y.​

Answer 1

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

Given :-

The vertices of parallelogram ABCD as

• • A (3, 3)

• • B (6, y)

• • C (x, 7)

• • D (5, 6)

Concept Used :-

We know,

In parallelogram, diagonals bisect each other.

So as it is given that ABCD is a parallelogram, it is sufficient to use the concept that midpoint of AC is equals to midpoint of BD to get the values of x and 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 AC.

• • Coordinates of A = ( 3, 3 )

• • Coordinates of C = ( x, 7 )

Here,

• • • x₁ = 3

• • • x₂ = x

• • • y₁ = 3

• • • y₂ = 7

So,

Using midpoint Formula,

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

$\rm :\longmapsto\: \sf \: Midpoint \: of \: AC \: = \: \bigg(\dfrac{3+x}{2} , \dfrac{10}{2} \bigg)$

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

Now, To find Midpoint of BD,

• • Coordinates of B = ( 6, y )

• • Coordinates of D = ( 5, 6 )

Here,

• • • x₁ = 6

• • • x₂ = 5

• • • y₁ = y

• • • y₂ = 6

So,

Using Midpoint Formula,

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

$\rm :\longmapsto\:Midpoint \: of \: BD = \bigg(\dfrac{6 + 5}{2} , \dfrac{y + 6}{2} \bigg)$

$\rm :\longmapsto\:Midpoint \: of \: BD = \bigg(\dfrac{11}{2} , \dfrac{y + 6}{2} \bigg)$

Now, Since ABCD is a parallelogram.

So,

Midpoint of AC = Midpoint of BD

$\red{\rm :\longmapsto\:\bigg(\dfrac{3+x}{2} , \: 5\bigg) = \bigg(\dfrac{11}{2} , \dfrac{y + 6}{2} \bigg)}$

On comparing,

$\rm :\longmapsto\:3 + x = 11\rm \implies\:x = 8$

and

$\rm :\longmapsto\:6 + y = 10\rm \implies\:y = 4$