Question

(5) If the points (-2, -1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the values of x and y.​

Answer 1

Answer :-

If the given coordinates are of a parallelogram taken in order then the mid points of both the diagonals will be same (the point where both diagonals meet).

Assume coordinates be A(-2,-1), B(1,0), C(x,3) and D(1,y).

Mid point of AC will be same of mid point of BD.

We have formula for mid-point::

$\sf\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)=\bigg(\dfrac{x_3+x_4}{2},\dfrac{y_3+y_4}{2}\bigg)$

Here::

» x₁ = -2

» x₂ = x

» x₃ = 1

» x₄ = 1

» y₁ = -1

» y₂ = 3

» y₃ = 0

» y₄ = y

Substitute these all values in above formula.

$\sf\bigg(\dfrac{-2+x}{2},\dfrac{-1+3}{2}\bigg)=\bigg(\dfrac{1+1}{2},\dfrac{0+y}{2}\bigg)$

$\sf\bigg(\dfrac{-2+x}{2},\dfrac{2}{2}\bigg)=\bigg(\dfrac{2}{2},\dfrac{y}{2}\bigg)$

$\sf\bigg(\dfrac{-2+x}{2},1\bigg)=\bigg(1,\dfrac{y}{2}\bigg)$

>> Now comparing LHS and RHS

$\sf\bigg(\dfrac{-2+x}{2}=1\bigg),\bigg(\dfrac{y}{2}=1\bigg)$

>> Now cross multiplication

$\sf({-2+x}=2),(y=2)$

$\sf({x}=2+2),(y=2)$

$\sf({x}=4),(y=2)$

This is the required values of x and y.

Answer 2

Answer:

Answer :-

If the given coordinates are of a parallelogram taken in order then the mid points of both the diagonals will be same (the point where both diagonals meet).

Assume coordinates be A(-2,-1), B(1,0), C(x,3) and D(1,y).

Mid point of AC will be same of mid point of BD.

We have formula for mid-point::

\sf\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)=\bigg(\dfrac{x_3+x_4}{2},\dfrac{y_3+y_4}{2}\bigg)(

2

x

1

+x

2

,

2

y

1

+y

2

)=(

2

x

3

+x

4

,

2

y

3

+y

4

)

Here::

» x₁ = -2

» x₂ = x

» x₃ = 1

» x₄ = 1

» y₁ = -1

» y₂ = 3

» y₃ = 0

» y₄ = y

Substitute these all values in above formula.

\sf\bigg(\dfrac{-2+x}{2},\dfrac{-1+3}{2}\bigg)=\bigg(\dfrac{1+1}{2},\dfrac{0+y}{2}\bigg)(

2

−2+x

,

2

−1+3

)=(

2

1+1

,

2

0+y

)

\sf\bigg(\dfrac{-2+x}{2},\dfrac{2}{2}\bigg)=\bigg(\dfrac{2}{2},\dfrac{y}{2}\bigg)(

2

−2+x

,

2

2

)=(

2

2

,

2

y

)

\sf\bigg(\dfrac{-2+x}{2},1\bigg)=\bigg(1,\dfrac{y}{2}\bigg)(

2

−2+x

,1)=(1,

2

y

)

>> Now comparing LHS and RHS

\sf\bigg(\dfrac{-2+x}{2}=1\bigg),\bigg(\dfrac{y}{2}=1\bigg)(

2

−2+x

=1),(

2

y

=1)

>> Now cross multiplication

\sf({-2+x}=2),(y=2)(−2+x=2),(y=2)

\sf({x}=2+2),(y=2)(x=2+2),(y=2)

\sf({x}=4),(y=2)(x=4),(y=2)

This is the required values of x and y.