Question

Three vertices of a parallelogram abcd taken an order are a 43 b - 1, 2 and 3 - 2, - 3 find the coordinates of the fourth vertex

Answer 1

Answer:

Given,

Three vertices of a parallelogram ABCD taken an order are A(4,3) ; B(- 1, 2) and C(- 2, - 3).

Let D( x,y) is the fourth vertex.

In a parallelogram ABCD diagonals are bisect each other then,

Coordinates of mid point of AC = Coordinates of mid point of BD

∴ using mid-point formula i.e. {(x₁ + x₂)/2 , (y₁ + y₂)/2}

we get,

∴ { (4 - 2 ) / 2 , (3 - 3) / 2 } = { (-1 + x ) / 2 , (2 + y) / 2 }

∴ (1 , 0 ) = { (-1 + x ) / 2 , (2 + y) / 2 }

∴ (-1 + x ) / 2 = 1 and (2 + y) / 2 = 0

∴ -1 + x = 2 and 2 + y = 0

∴ x = 3 and y = -2

∴The fourth vertex = ( 3 ,-2)

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

using mid-point formula i.e. {(x₁ + x₂)/2 , (y₁ + y₂)/2}

∴ { (4 - 2 ) / 2 , (3 - 3) / 2 } = { (-1 + x ) / 2 , (2 + y) / 2 }

∴ (1 , 0 ) =  (-1 + x ) / 2 , (2 + y) / 2

∴ (-1 + x ) / 2 = 1 and (2 + y) / 2 = 0

∴ -1 + x = 2 and 2 + y = 0

∴ x = 3 and y = -2