Question

If (1,2), (3,0), (a,4), (5,b) are the vertices of parallelogram, find the value of a and b​

Answer 1

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

Given that the vertices of parallelogram are (1,2), (3,0), (a,4) and (5,b).

Let us assume the vertices as A, B, C, D such that

• A (1,2)

• B (3,0)

• C (a,4)

• D (5,b)

Concept Used :-

We know,

In parallelogram, diagonals bisect each other.

So in order to find the values of a and b from the given vertices of A, B, C, D taken in order forms a parallelogram, we use the concept that midpoint of AC is equals to midpoint of BD.

Now,

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

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

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

Now,

Let us first find midpoint of AC.

• Coordinates of A = ( 1, 2)

• Coordinates of C = (a, 4)

Using midpoint Formula,

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

Here,

• x₁ = 1

• x₂ = a

• y₁ = 2

• y₂ = 4

So,

$\rm :\longmapsto\:{ \quad\sf \:Midpoint \: of \: AC= \bigg(\dfrac{1+a}{2} , \dfrac{2+4}{2} \bigg) \quad}$

$\rm :\longmapsto\:{ \quad\sf \:Midpoint \: of \: AC= \bigg(\dfrac{1+a}{2} , \dfrac{6}{2} \bigg) \quad}$

$\boxed{ \quad\sf \:Midpoint \: of \: AC= \bigg(\dfrac{1+a}{2} , \: 3 \bigg) \quad}$

Now,

Lets find midpoint of BD.

• Coordinates of A = ( 3, 0)

• Coordinates of C = (5, b)

Using midpoint Formula,

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

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

$\rm :\longmapsto\:{ \quad\sf \:Midpoint \: of \:BD = \bigg(\dfrac{8}{2} , \dfrac{b}{2} \bigg) \quad}$

$\boxed{ \quad\sf \:Midpoint \: of \:BD = \bigg(4 , \: \dfrac{b}{2} \bigg) \quad}$

Now, we have

Midpoint of AC = Midpoint of BD

So,

$\boxed{ \quad\sf \: \bigg( \dfrac{a + 1}{2}, \:3 \bigg) \: = \: \bigg(4 , \: \dfrac{b}{2} \bigg) \quad}$

So, on comparing, we get

$\rm :\longmapsto\:\dfrac{a + 1}{2} = 4$

$\rm :\longmapsto\:a + 1 = 8$

$\bf\implies \:a = 7$

Also,

$\rm :\longmapsto\:\dfrac{b}{2} = 3$

$\bf\implies \:b = 6$