Question

If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.
OR
If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Answer 1

Given, ABCD is a parallelogram.

Midpoint of AC $=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$=(\frac{-2+4}{2},\frac{1+b}{2})$
$=(\frac{2}{2},\frac{1+b}{2})$
$=(1,\frac{1+b}{2})$
Midpoint of BD = $(\frac{x_1^{\prime }+x_2^{\prime }}{2},\frac{y_1^{\prime }+y_2^{\prime }}{2})$
= $(\frac{a+1}{2},\frac{0+2}{2})$
= $(\frac{a+1}{2},\frac{2}{2})$
= $(\frac{a+1}{2},1)$
Since, diagonals of a parallelogram bisect each ether,
$\therefore (1,\frac{1+b}{2})=(\frac{a+1}{2},1)$
On comparing, we get
$\therefore \frac{a+1}{2}=1$ | $\frac{1+b}{2}=1$
$\Rightarrow a+1=2$ | $\Rightarrow 1+b=2$
$\Rightarrow a=1$ | $\Rightarrow$ b = 1
Therefore, the coordinates of vertices of parallelogram ABCD are A(-2, 1), B(1, 0), C(4, 1) and D(1, 2)
Length of side AB = DC = $\sqrt{(1+2{)}^2+(0-1{)}^2}$
= $\sqrt{9+1}=\sqrt{10}$ units
And, AD = BC = $\sqrt{(1+2{)}^2+(2-1{)}^2}$
$=\sqrt{9+1}=\sqrt{10}$ units

OR


Given ABCD is quadrilateral.
By joining points A and C, the quadrilateral is divided into two triangles.
Now, Area of quad. ABCD = Area of $\Delta ABC$ + Area of $\Delta ACD$
Area of $\Delta ABC$
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
$=\frac{1}{2}[-5(-5+6)-4(-6-7)-1(7+5\left)\right]$
$=\frac{1}{2}[-5(1)-4(-13)-1(12\left)\right]$
$=\frac{1}{2}(-5+52-12)$
$-\frac{1}{2}\left(35\right)=\frac{35}{2}$sq.units.
Area of $\Delta$ ADC
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
$=\frac{1}{2}[-5(5+6)+4(-6-7)+(-1\left)\right(7-5\left)\right]$
$=\frac{1}{2}[-5(11)+4(-13)-1(2\left)\right]$
$=\frac{1}{2}(-55-52-2)$
$=\frac{1}{2}|-109|=\frac{109}{2}$ sq. units.
Area of quadrilateral ABCD
$=\frac{35}{2}+\frac{109}{2}$
$=\frac{144}{2}=72$ sq. units.