Question

The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.

Answer 1

Let A(3,4), B(3,8) and C(9,8) be the three given vertex then the fourth vertex D(x,y)

Since ABCD is a parallelogram, the diagonals bisect each other.

Therefore the mid-point of diagonals of the parallelogram coincide.

Let p(x,y) be the mid-point of diagonals AC then,

${\displaystyle P(x,y)=(\frac{3+9}{2},\frac{4+8}{2})}$

P(x,y) = (6,6)

Let Q(x,y) be the mid point of diagonal BD. then

${\displaystyle Q(x,y)=(\frac{3+x}{2},\frac{8+y}{2})}$

Coordinates of mid-point AC = Coordinates of mid-point BD

P(x,y) = Q(x,y)

${\displaystyle \Rightarrow (6,6)=(\frac{3+x}{2},\frac{8+y}{2})}$

Now equating individual components

${\displaystyle \Rightarrow 6=\frac{3+x}{2}}$ and ${\displaystyle 6=\frac{8+4}{2}}$

=> 3 + x = 12 and 8 + y = 12

=> x = 9 and y = 4

Hence, coordinates of fourth points are (9, 4)