Question

if two adjacent vertices of a parallelogram are (-1,0) (3,5) and the diagonals cut at (2,4) , find the vertices of a parallelogram​

Answer 1

Answer:

C = (5 , 8)

D = (1 , 3)

Step-by-step explanation:

Given,

A = (-1 , 0)

B = (3 , 5)

Diagonals intersect at(O) =  (2 , 4)

Let , the parallelogram be ABCD

To Find :-

Co-ordinates of Vertices of parallelogram(C , D)

How To Do :-

As they said the intersecting point of diagonals(O) , In a parallelogram intersecting point of both diagonals is the mid - point of diagonals AC , BD.

So, y using the Mid-point formula We need to find the value of remaining vertices C , D.

Formula Required :-

Mid - point formula :-

$(x,y)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Solution :-

∴ 'O' is the Mid-point of both AC , BD

Taking For AC :-

O = (2 , 4)

x = 2 , y = 4

A = (-1 , 0)

x_1 = -1 , y_1 = 0

Let, C = (x_2 , y_2)

Substituting in Mid-point formula :-

$(2,4)=\left(\dfrac{-1+x_2}{2},\dfrac{0+y_2}{2}\right)$

Equating both 'x' terms and 'y' terms :

2 = - 1 + x_2 /2 , 4 = 0 + y_2/2

First taking x-terms :-

2 = -1+x_2/2

2(2) = - 1 + x_2

4 = - 1 + x_2

x_2 = 4 + 1

x_2 = 5

Taking 'y terms' :-

4 = 0 + y_2/2

4(2) = 0 + y_2

8 = 0 + y_2

y_2 = 8

∴ Co-ordinates of 'C' = (5 , 8)

Taking For BD :-

O = (2 , 4)

x = 2 , y = 4

B = (3 , 5)

x_1 = 3 , y_1 = 5

Let, D = (x_2 , y_2)

Substituting in Mid-point formula :-

$(2,4)=\left(\dfrac{3+x_2}{2},\dfrac{5+y_2}{2}\right)$

Equating both 'x' terms and 'y' terms :-

First equating 'x' term :-

2 = 3 + x_2/2

2(2) = 3 + x_2

4 = 3 + x_2

x_2 =  4 - 3

x_2 = 1

Equating 'y' terms :-

4 = 5 + y_2/2

4(2) = 5 + y_2

8 = 5 + y_2

y_2 = 8 - 5

y_2 = 3

∴ Co-ordinates of 'D' = (1 , 3)