Question

Three vertices of a parallelogram are shown in the figure below.

Answer 1

Given:

The three vertices of the parallelogram are (-3,9), (0,-3), (6,-6).

To find:

The fourth vertex of the parallelogram.

Solution:

Consider the three vertices of the parallelogram are A(-3,9), B(0,-3), C(6,-6).

Let D(a,b) be the fourth vertex.

Midpoint formula:

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

We know that the diagonals of a parallelogram bisect each other. So, the midpoints of both diagonals are same.

Midpoint of AC = Midpoint BD

$\left(\dfrac{-3+6}{2},\dfrac{9+(-6)}{2}\right)=\left(\dfrac{0+a}{2},\dfrac{-3+b}{2}\right)$

$\left(\dfrac{3}{2},\dfrac{3}{2}\right)=\left(\dfrac{a}{2},\dfrac{-3+b}{2}\right)$

On comparing both sides, we get

$\dfrac{3}{2}=\dfrac{a}{2}$

$3=a$

And,

$\dfrac{3}{2}=\dfrac{-3+b}{2}$

$3=-3+b$

$3+3=b$

$6=b$

Therefore, the fourth vertex of the parallelogram is (3,6).