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

Answer: The co-ordinates of the vertex D are (-2, 1). and the area of ABCD is 15 sq. units.

Step-by-step explanation: As given in the question and shown in the attached figure, ABCD is a parallelogram, where the three vertices are A(3,-4) B(-1,-3) and C(-6,2).

We are to find the co-ordinates of the vertex 'D'. Let (a,b) be the co-ordinates of the vertex 'D'.

Since the opposite sides of a parallelogram are parallel, so their slopes must be equal.

In the parallelogram ABCD, we have

$\textup{Slope of AB}=\textup{Slope of CD}\\\\\Rightarrow \dfrac{-3+4}{-1-3}=\dfrac{b-2}{a+6}\\\\\Rightarrow \dfrac{1}{-4}=\dfrac{b-2}{a+6}\\\\\Rightarrow a+6=-4b+8\\\\\Rightarrow a=-4b+2,~~~~~~~~~~~~~~~~(i)$

and

$\textup{Slope of AD}=\textup{Slope of BC}\\\\\Rightarrow \dfrac{b+4}{a-3}=\dfrac{2+3}{-6+1}\\\\\Rightarrow \dfrac{b+4}{a-3}=\dfrac{5}{-5}\\\\\Rightarrow a-3=-b-4\\\\\Rightarrow a=-b-1,~~~~~~~~~~~~~~~~(i)$

Comparing equations (i) and (ii), we have

$-4b+2=-b-1\\\\\Rightarrow -3b=-3\\\\\Rightarrow b=1,$

and from equation (ii), we get

$a=-1-1=-2.$

Thus, the co-ordinates of the vertex D are (-2, 1).

Now, let us divide the parallelogram ABCD in two congruent triangles ACD and ABD since a diagonal divides parallelogram into two congruent triangles.

Area of ΔABD is given by

$\triangle=\dfrac{1}{2}|3(-3-2)-1(2+4)-6(-4+3)|=\dfrac{1}{2}|-15-6+6|=7.5~\textup{sq. units}.$

Thus, the area of ABCD will be

$Area_{ABCD}=2\times\triangle=2\times7.5=15~\textup{sq. units}.$