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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.
Given:
The points $A (1, -2), B (2, 3), C (a, 2)$ and $D (-4, -3)$ form a parallelogram.
To do:
We have to find the value of $a$ and height of the parallelogram taking $AB$ as base.
Solution:
Draw a perpendicular from \( \mathrm{D} \) to \( \mathrm{AB} \) which meets \( \mathrm{AB} \) at \( \mathrm{P} \).
\( \mathrm{DP} \) is the height of the parallelogram.
We know that,
Diagonals of a parallelogram bisect each other.
This implies,
Mid-point of $AC =$ Mid-point of $BD$
The mid-point of a line segment joining points \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \) is \( (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \)
\( (\frac{1+a}{2}, \frac{-2+2}{2})=(\frac{2-4}{2}, \frac{3-3}{2}) \)
On comparing, we get,
\( \Rightarrow \frac{1+a}{2}=\frac{-2}{2}=-1 \)
\( \Rightarrow 1+a=-2 \)
\( \Rightarrow a=-3 \)
The required value of \( a \) is \( -3 \).
We know that,
A diagonal bisects a triangle into two triangles of equal area.
This implies,
Area of parallelogram $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
$=2\times$ Area of triangle $ABC$
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[1(3-2)+2(2+2)+(-3)(-2-3)] \)
\( =\frac{1}{2}[1(1)+(2)(4)+(-3)(-5)] \)
\( =\frac{1}{2}[1+8+15] \)
\( =\frac{1}{2} \times (24) \)
\( =12 \) sq. units.
Therefore,
The area of the parallelogram $ABCD=2\times 12=24$ sq. units.
Area of parallelogram $=$ Base $\times$ Height
Height $=$ Area $\div$ Base
$DP=\frac{24}{AB}$
By distance formula, we get,
$AB=\sqrt{(2-1)^2+(3+2)^2}$
$=\sqrt{1^2+5^2}$
$=\sqrt{1+25}$
$=\sqrt{26}$
Therefore,
$DP=\frac{24}{\sqrt{26}}$
$=\frac{24\times\sqrt{26}}{\sqrt{26}\times\sqrt{26}}$
$=\frac{24\sqrt{26}}{26}$
$=\frac{12\sqrt{26}}{13}$
The height of the parallelogram taking $AB$ as base is $\frac{12\sqrt{26}}{13}$.
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