Given, the point A(1, -2), B(2, 3), C(-3, 2) and D(-4, -3) form a parallelogram.
The area of a triangle with vertices \(A\left(x_1, y_1\right), B\left(x_2, y_2\right)\) and \(c\left(x_3, y_3\right)\) is.
\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)
Here, \(\left(x_1, y_1\right)=(1,-2),\left(x_2, y_2\right)=(2,3)\) and \(\left(x_3, y_3\right)=(-3,2)\)
Area of \(\triangle ABC = \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)
\(=\frac{1}{2}[1(3-2)+2(2-(-2))+(-3)(-2-3)] \)
\(=\frac{1}{2}[1(1)+2(2+2)-3(-5)] \)
\(=\frac{1}{2}[1+8+15] \)
\(=\frac{1}{2} \times 24\)
= 12 square units.
Now, the area of parallelogram ABCD =2 x area of triangle ABC
\(=2 \times 12\)
= 24 square units.
Base = AB
The distance between two points \(P\left(x_1, y_1\right)\) and \(Q\left(x_2, y_2\right)\) is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Distance between \(A(1,-2)\) and \(B(2,3)\)
\(=\sqrt{(2-1)^2+(3-(-2))^2} \)
\(=\sqrt{(1)^2+(3+2)^2} \)
\(=\sqrt{(1)^2+(5)^2} \)
\(=\sqrt{1+25} \)
\(=\sqrt{26}\)
We know that area of parallelogram = base x height
\(24 =\sqrt{26} \times \text { height } \)
\(\text { Height } =\frac{24}{\sqrt{26}}\)
Therefore, the height of the parallelogram is \(\frac{24}{\sqrt{26}}\) units.
