Question

The three vertices of a parallelogram ABCD are A(3, −4), B(−1, −3) and C(−6, 2). Find the coordinates of vertex D and find the area of ABCD.

Answer 1

The three vertices of the parallelogram ABCD are A (3, −4), B (−1, −3) and C (−6, 2).

Let the coordinates of the vertex D be (x, y).

It is known that in a parallelogram, the diagonals bisect each other.

∴Mid point of AC = Mid point of BD

`rArr ((3-6)/2,(-4+2)/2)=((-1+x)/2,(-3+y)/2)`

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

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

`rArrx=-2,y=1`

So, the coordinates of the vertex D is (−2, 1).

Now, area of parallelogram ABCD

= area of triangle ABC + area of triangle ACD

= 2 × area of triangle ABC [Diagonal divides the parallelogram into two triangles of equal area]

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by the numerical value of the expression 

`1/2[x_1(y_2-y_3)+x_2(y_3-y_1+x_3(y_1-y_2)]`

Area of triangle ABC =`1/2[3(-3-2)+(-1){2-(-4)}+(-6){-4-(3)}]`

`rArr1/2[3xx (-5)+(-1)xx6+(-6)xx(-1)]=1/2[-15-6+6]=-15/2`

∴Area of triangle ABC = `15/2`square units (Area of the triangle cannot be negative)

Thus, the area of parallelogram ABCD `=2xx15/2=15`square units.