Question

If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a
parallelogram PQRS, find the values of a and b.​

Answer 1

Answer:

If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a

parallelogram PQRS, find the values of a and b.

Answer 2

Given :

The vertices of a parallelogram PQRS,

• P(a, -11)
• Q(5, b)
• R(2, 15)
• S(1, 1)

To Find :

The values of a and b.

Solution :

Analysis :

Here the concept of mid-formula is used. Here we are given with all the four vertices of a parallelogram. We know that the diagonals of a parallelogram bisect each other. According to this property, the mid-points of PR and QS will be equal. Now by equating we will get the values of a and b.

Explanation :

Since the diagonals bisect each other, the mid-points of PR and QS will be equal.

Mid-Point of PR :

• P(a, -11)
• R(2, 15)

$\boxed{\pmb{\sf\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}$

where,

• x₁ = a
• x₂ = 2
• y₁ = -11
• y₂ = 15

$\implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{-11+15}{2}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{4}{2}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{\not{4}}{\not{2}}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},2\bigg\rgroup$

$\therefore$ Mid-Point of PR is $\pmb{\sf\bigg\lgroup\dfrac{a+2}{2},2\bigg\rgroup}$ (eq.(i))

Now,

Mid-Point of QS :

• Q(5, b)
• S(1, 1)

$\boxed{\pmb{\sf\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}$

where,

• x₁ = 5
• x₂ = 1
• y₁ = b
• y₂ = 1

$\implies\sf\bigg\lgroup\dfrac{5+1}{2},\dfrac{b+1}{2}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup\dfrac{6}{2},\dfrac{b+1}{2}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup\dfrac{\not{6}}{\not{2}},\dfrac{b+1}{2}\bigg\rgroup$

$\\ \implies\sf\bigg\lgroup3,\dfrac{b+1}{2}\bigg\rgroup$

$\therefore$ Mid-Point of QS is $\pmb{\sf\bigg\lgroup 3,\dfrac{b+1}{2}\bigg\rgroup}$ (eq.(ii))

• Mid-Points of PR and QS are same.

From eq.(i) and eq.(ii),

$\implies\sf\dfrac{a+2}{2}=3$

$\implies\sf a+2=3\times2$

$\implies\sf a+2=6$

$\implies\sf a=6-2$

$\therefore\boxed{\pmb{\sf a=4.}}$

Again,

From eq.(i) and eq.(ii),

$\implies\sf\dfrac{b+1}{2}=2$

$\implies\sf b+1=2\times2$

$\implies\sf b+1=4$

$\implies\sf b=4-1$

$\therefore\boxed{\pmb{\sf b=3.}}$

So, the values of a and b are,

$\boxed{\begin{cases}&\bf{a=4} \\ &\bf{b=3}\end{cases}}$