Question

If the points $A(1,-2),B(2,3),C(a,2)\mathrm{and}D(-4,-3)$ 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 eachother.
Coordinates of the midpoint of AC = coordinates of the midpoint of BD.
$$\begin{gathered} \Rightarrow \left(\frac{a+1}{2},\frac{2-2}{2}\right)=\left(\frac{-4+2}{2},\frac{-3+3}{2}\right) \\ \Rightarrow \left(\frac{a+1}{2},0\right)=\left(-1,0\right) \\ \mathrm{On}\mathrm{comparing}, \\ \Rightarrow \frac{a+1}{2}=-1 \\ \Rightarrow a=-3 \\ \mathrm{Area}\mathrm{of}\mathrm{the}∆\mathrm{ABC}\mathrm{is} \\ A=\frac{1}{2}\left[1\left(3-2\right)+2\left(2+2\right)-3\left(-2-3\right)\right] \\ A=\frac{1}{2}\left[1+8+15\right] \\ A=12\mathrm{sq}.\mathrm{units} \end{gathered}$$

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