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
$(\frac{3-6}{2},\frac{-4+2}{2})=(\frac{-1+x}{2},\frac{-3+y}{2})(-\frac{3}{2},-\frac{2}{2})=(\frac{-1+x}{2},\frac{-3+y}{2})-\frac{3}{2}=\frac{-1+x}{2},-\frac{2}{2}=\frac{-3+y}{2}x=-2,y=1$
So, the coordinates of the vertex x D is (-2, 1).
Now, area of parallelogram ABCD
= area of triangle ABC + area of triangle ACD
$=2\times$ area of triangle ABC [Diagonal divides the parallelogram into two triangles of equal area]
The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is given by the numerical value of the expression $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
Area of triangle $ABC=\frac{1}{2}\left[3\right(-3-2)+(-1\left)\right\{2-(-4)\}+(-6\left)\right\{-4-(-3)\left\}\right]$
$=\frac{1}{2}[3\times (-5)+(-1)\times 6+(-6)\times (-1\left)\right]=\frac{1}{2}[-15-6+6]=-\frac{15}{2}$
$\therefore$ Area of triangle $ABC=\frac{15}{2}$ sq units (Area of the triangle cannot be negative)
Thus, the area of parallelogram $ABCD=2\times \frac{15}{2}=15$ sq units.