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.

$$\Rightarrow (\frac{a+1}{2},\frac{2-2}{2})=(\frac{-4+2}{2},\frac{-3+3}{2})$$

$$\Rightarrow (\frac{a+1}{2},0)=(-1,0)$$

$$\text{ On comparing, }$$

$$\Rightarrow \frac{a+1}{2}=-1$$

$$\Rightarrow a=-3$$

$$\text{ Area of the ∆ ABC is }$$

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

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

$$A=12\text{ sq . units }$$

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 $$AB=\sqrt{{(1-2)}^2+{(-2-3)}^2}$$
$$=\sqrt{1^2+5^2}$$
$$=\sqrt{26}$$
$$\begin{gathered} \text{ Height }=\frac{24}{\sqrt{26}}=\frac{12\sqrt{26}}{ \\ 13} \end{gathered}$$