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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.

By Margaret

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.​

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Margaret

The perimeter of a rectangular field is 100 m. If length of the
field is 35 m, find its area.

Please tell me this

F. Fill in the blanks:
1. The land use is directly linked with__________________
activities.

2. Land use also

2 thoughts on “If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a<br />parallelogram PQRS, find the values of a and b.”

  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.

    Reply

  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)

    [tex]\boxed{\pmb{\sf\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}[/tex]

    where,

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

    [tex]\implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{-11+15}{2}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{4}{2}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},\dfrac{\not{4}}{\not{2}}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup\dfrac{a+2}{2},2\bigg\rgroup[/tex]

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

    Now,

    Mid-Point of QS :

    • Q(5, b)
    • S(1, 1)

    [tex]\boxed{\pmb{\sf\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}[/tex]

    where,

    • x₁ = 5
    • x₂ = 1
    • y₁ = b
    • y₂ = 1

    [tex]\implies\sf\bigg\lgroup\dfrac{5+1}{2},\dfrac{b+1}{2}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup\dfrac{6}{2},\dfrac{b+1}{2}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup\dfrac{\not{6}}{\not{2}},\dfrac{b+1}{2}\bigg\rgroup[/tex]

    [tex]\\ \implies\sf\bigg\lgroup3,\dfrac{b+1}{2}\bigg\rgroup[/tex]

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

    • Mid-Points of PR and QS are same.

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

    [tex]\implies\sf\dfrac{a+2}{2}=3[/tex]

    [tex]\implies\sf a+2=3\times2[/tex]

    [tex]\implies\sf a+2=6[/tex]

    [tex]\implies\sf a=6-2[/tex]

    [tex]\therefore\boxed{\pmb{\sf a=4.}}[/tex]

    Again,

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

    [tex]\implies\sf\dfrac{b+1}{2}=2[/tex]

    [tex]\implies\sf b+1=2\times2[/tex]

    [tex]\implies\sf b+1=4[/tex]

    [tex]\implies\sf b=4-1[/tex]

    [tex]\therefore\boxed{\pmb{\sf b=3.}}[/tex]

    [tex]\\ [/tex]

    So, the values of a and b are,

    [tex]\boxed{\begin{cases}&\bf{a=4} \\ &\bf{b=3}\end{cases}}[/tex]

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