Question

$\huge \mathfrak \blue{Question}$

❥ 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 value of a and b.

[ Class 10 ]
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Answer 1

Answer:

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➜Let A(a, -11), B(5, b), C(2, 15) and D(1,1) be the given points.

★We know that diagonals of parallelogram bisect each other

★Therefore, Coordinates of mid-point of AC = Coordinates of mid-point of BD

$( \frac{a + 2}{2} , \frac{15 - 11}{2} ) = ( \frac{5 + 1}{2} , \frac{b + 1}{2} )$

$➜ \frac{a + 2}{2} = 3 \: and \: \frac{b + 1}{2} = 2$

$➜a + 2 = 6 \: ➜b + 1 = 4$

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$➜a = 4 \: ➜ \: b = 3$

Answer 2

Given :

The points P (a, –11), Q (5, b), R (2, 15) and S (1, 1) are the vertices of a parallelogram PQRS.

To Find :

The value of a and b.

Solution :

We are know about the properties of parallelogram.

• The diagonals of parallelogram bisects each other.
• Each diagonal of a parallelogram separates it into two congruent triangles.

Now,

Coordinate of mid point of AC = Coordinate of mid point of BD.

$\sf \bigg[ \dfrac {a + 2}{2} , \ \dfrac {-11 + 15}{2} \bigg] \ = \ \bigg[ \dfrac {5+1}{2}, \ \dfrac {b+1}{2} \bigg]$

$\implies \sf \dfrac {a+2}{2} \ = \ 3 \ and \ \dfrac {b+1}{2} \ = \ 2$

$\implies \sf a \ + \ 2 \ = \ 6 \ and \ b \ + \ 1 \ = \ 4$

$\implies \sf a \ = \ 6 - 2 \ and \ b \ = \ 4 - 1$

$\implies \sf a \ = \ 4 \ and \ b \ = \ 3$

Hence :

• The value of a is 4.
• The value of b is 3.