Question

If the points A(1, –2), B(2, 3) C(a, 2) and D(– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.  

Answer 1

Since diagonals of a parallelogram bisect each other.
Coordinates of the midpoint of AC = coordinates of the midpoint of BD.

\[\Rightarrow \left( \frac{a + 1}{2}, \frac{2 - 2}{2} \right) = \left( \frac{- 4 + 2}{2}, \frac{- 3 + 3}{2} \right)\]

\[ \Rightarrow \left( \frac{a + 1}{2}, 0 \right) = \left( - 1, 0 \right)\]

\[\text{ On comparing, }  \]

\[ \Rightarrow \frac{a + 1}{2} = - 1\]

\[ \Rightarrow a = - 3\]

\[\text{ Area of the ∆ ABC is } \]

\[A = \frac{1}{2}\left[ 1\left( 3 - 2 \right) + 2\left( 2 + 2 \right) - 3\left( - 2 - 3 \right) \right]\]

\[A = \frac{1}{2}\left[ 1 + 8 + 15 \right]\]

\[A = 12 \text{ sq . units } \]

Since, ABCD is a parallelogram,
Area of ABCD = 2 × area of triangle ABC  = 2 × 12 = 24 sq. units
Height of the parallelogram is area of the parallelogram divided by its base.
Base AB is \[AB = \sqrt{\left( 1 - 2 \right)^2 + \left( - 2 - 3 \right)^2}\]
\[ = \sqrt{1^2 + 5^2}\]
\[ = \sqrt{26}\]
\[\text{ Height }  = \frac{24}{\sqrt{26}} = \frac{12\sqrt{26}}{\\{13}}\]