Question

Find the area of a parallelogram given these vertices: P1(1,2) P2(4,4) P3(7,5) P4(4,3)?

Answer 1

Area of parallelogram is 3

Explanation:

Given A (1,2), B (4,4), C(3) 7,5), D (4,3)

`AB = l = sqrt((4-1)^2 + (4-2)^2) = 3.6056`

`AD = w = sqrt((4-1)^2 + (3-2)^2) = 3.1623`

`BC = w = sqrt((7-4)^2 + (5-4)^2)) = 3.1632`

`CD = l = sqrt((7-4)^2 + (5-3)^2) = 3.6056`

Slope of `AB = m = (4-2) / (4-1) = (2/3)`

Eqn of AB
`(y-2) / (4-2) = (x-1) / (4-1)`
`3y - 6 = 2x - 2`
`2x - 3y = -4` Eqn (1)

Slope of `DE = m_1 = = -(1/m) = -(3/2)`

Eqn of DE is
`(y-3) = -(3/2)*(x-4)`
`2y - 6 = -3x + 12`
`3x + 2y = 18`. Eqn (2)

Solving equations (1) & (2) we get coordinates of point E.
Coordinates of `E(46/13, 48/13)`

Length of `DE = h = sqrt((4-(46/13)^2 + (3-(48/13)^2) 0.8321`

Area of parallelogram ABCD = `l * h = 3.6056 * 0.8321 = color (purple)3`

Answer 2

Contd.....

Explanation:

Prerequisites : #Area of the Delta with vertices (x_1,y_1), (x_2,y_2) and (x_3,y_3) is 1/2|D|, where, D=|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|#.