Find the area of the parallelogram whose three of the vertices are (1, -2), (2, 3) and (-3, 2) in order. Also find its fourth vertex .

Question

asked Jul 8, 2020 63.9k views

2 Answers

Rick Arthur · Answer 1 · 2020-07-12T22:01:03+0000

Answer:

Area = 24 square unit,

Fourth vertex = (-4, -3)

Explanation:

Suppose we have a parallelogram ABCD,

Having vertex A(1, -2), B(2, 3), and C(-3, 2),

Let D(x,y) be the fourth vertex of the parallelogram,

∵ The diagonals of a parallelogram bisect each other,

Thus, the midpoint of AC = midpoint of BD

((1-3)/(2), (-2+2)/(2))=((2+x)/(2), (3+y)/(2))

((-2)/(2), 0)=((2+x)/(2), (3+y)/(2))

By comparing,

$-2=2+x\implies x=-4$

$3+y=0\implies y = -3$

Thus, the fourth vertex is (-4, -3),

Now, the area of the parallelogram ABCD = 2 × area of triangle ABC (Because both diagonals divide the parallelogram in two equal triangles)

Area of a triangle having vertex
(x_1, y_1) ,
(x_2, y_2) and
(x_3, y_3) is,

A=(1)/(2)|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

So, the area of triangle ABC

A=(1)/(2)|(1(3-2)+2(2+2)-3(-2-3)}|

=(1)/(2)(1+8+15)

=(1)/(2)* 24

$=12\text{ square unit}$

Hence, the area of the parallelogram ABCD = 2 × 12 = 24 square unit.