Question

If the points $A(1,-2),B(2,3),C(a,2)$ and $D(-4,-3)$ from a parallelogram. Find the value of ${}^{\prime }{a}^{\prime }$ and height of the parallelogram taking $AB$ as base.

Answer 1

Consider the given points $A(1,-2),B(2,3),C(0,2)$ and $D(-4,-3)$

Since ABCD form a parallelogram, the midpoint of the diagonal AC should coincide with the midpoint of BD.

Mid point of AC$=$ Mid point of BD

$[\frac{1+a}{2},\frac{-2+2}{2}]=[\frac{2-4}{2},\frac{3-3}{2}]$

$[\frac{a+1}{2},0]=[\frac{-2}{2},0]$

Since the mid points coincide, we have

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

$\Rightarrow a+1=-2$

$\Rightarrow a=-2-1$

$\Rightarrow a=-3$

Now, area of $\Delta ABC$

$=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

$=\frac{1}{2}\left|1\right(3-2)+2(2-(-2))+(-3\left)\right(-2-3\left)\right|$

$=\frac{1}{2}\left|1\right(1)+2(4)+(-3\left)\right(-5\left)\right|$

$=\frac{1}{2}|1+8+15|$

$=\frac{24}{2}=12$ sq. units

$ar\left(ABCD\right)$ parallelogram $=2\times$ Area of triangle

$=2\times 12$

$=24$ sq. units

Area of parallelogram $=$Base $\times$ Height

$\frac{Area}{Base}=height$

So by the distance formula

$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$=\sqrt{(-3+4{)}^2+(2+3{)}^2}$

$=\sqrt{1+25}$

$=\sqrt{26}$

Thus height $=\frac{24}{\sqrt{26}}=\frac{24}{\sqrt{26}}\times \frac{\sqrt{26}}{\sqrt{26}}=\frac{24\sqrt{26}}{26}=\frac{12\sqrt{26}}{13}$.