Question

Three consecutive vertices
of a parallelogram, taken in
order are (6,8) (3,7), (-2,-2)
then the fourth vertex is​

Answer 1

Solution :-

Let the vertices of the parallelogram be A (6 , 8), B (3 , 7), C (-2 , -2) and D (a , b).

We know that, in a parallelogram diagonals bisect each other.

That is,

Midpoint of AC = Midpoint of BD

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

$\bullet \sf \ Midpoint \ of \ AC = \left( \dfrac{6+-2}{2} \ , \ \dfrac{8+-2}{2} \right)$

                           $= \sf \left( \dfrac{4}{2} \ , \ \dfrac{6}{2} \right) \\\\= ( 2 \ , 3 )$

$\bullet \sf \ Midpoint \ of \ BD = \left( \dfrac{3+a}{2} \ , \ \dfrac{7+b}{2} \right)$

$\longrightarrow \sf \dfrac{3+a}{2} =2 \\\\\longrightarrow 3+a = 4 \\\\\longrightarrow\bf a = 1$

$\longrightarrow \sf \dfrac{7+b}{2} = 3 \\\\\longrightarrow 7+b = 6 \\\\\longrightarrow\bf b = -1$

Coordinates of the fourth vertex (D) = ( 1 , -1 )

Answer 2

Step-by-step explanation:

Solution :-

Let the vertices of the parallelogram be A (6 , 8), B (3 , 7), C (-2 , -2) and D (a , b).

We know that, in a parallelogram diagonals bisect each other.

That is,

Midpoint of AC = Midpoint of BD

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