Question

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 .

Answer 1

do it

like this

i have done by coordinates of geometry

Answer 2

Answer:

Area = 24 square unit,

Fourth vertex = (-4, -3)

Step-by-step 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

$(\frac{1-3}{2}, \frac{-2+2}{2})=(\frac{2+x}{2}, \frac{3+y}{2})$

$(\frac{-2}{2}, 0)=(\frac{2+x}{2}, \frac{3+y}{2})$

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,

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

So, the area of triangle ABC

$A=\frac{1}{2}|(1(3-2)+2(2+2)-3(-2-3)}|$

$=\frac{1}{2}(1+8+15)$

$=\frac{1}{2}\times 24$

$=12\text{ square unit}$

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