Question

Question 2
A(6,1) B(8,2) and C(9,4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, then find the area of $\Delta$ ADE.

Answer 1

Given that, A(6,1) B(8,2) and C(9,4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram, be (x,y)
We know that, the diagonals of a parallelogram bisect each other.

$\therefore$ Mid-point of BD = Mid-point of AC
$\Rightarrow (\frac{8+x}{2},\frac{2+y}{2})=(\frac{6+9}{2},\frac{1+4}{2})[\because \text{Mid-point of a line segment joining the points}(x_1,y_1)\text{and}(x_2,y_2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})]\Rightarrow (\frac{8+x}{2},\frac{2+y}{2})=(\frac{15}{2},\frac{5}{2})\therefore \frac{8+x}{2}=\frac{15}{2}\Rightarrow 8+x=15\Rightarrow x=7And\frac{2+y}{2}=\frac{5}{2}\Rightarrow 2+y=5\Rightarrow y=3$
So, fourth vertex of a parallelogram is D(7,3).
Now, mid-point of side
$DC=(\frac{7+9}{2},\frac{3+4}{2})$
$\Rightarrow E=(8,\frac{7}{2})$
[$\because$ $\text{Area of}$ $\Delta ABC$ with vertices
$(x_1,y_1)$,$(x_2,y_2)\text{and}(x_3,y_3)$
$=\frac{1}{2}$$\left[x_1\right(y_2-y_3)$+$x_2(y_3-y_1)$+$x_3(y_1-y_2)]$
$\therefore \text{Area of}\Delta \text{ADE with vertices}$
$A(6,1),D(7,3)\text{and}E(8,\frac{7}{2})$
$\Delta =\frac{1}{2}$$\left[6\right(3-\frac{7}{2})$+$7(\frac{7}{2}-1)+8(1-3)]$
$=\frac{1}{2}$$[6\times (\frac{-1}{2})+7(\frac{5}{2})+8(-2\left)\right)\left)\right]$
$=\frac{1}{2}(-3+\frac{35}{2}-16)$
$=\frac{1}{2}(\frac{35}{2}-19)=\frac{1}{2}\left(\frac{-3}{2}\right)$
$=\frac{-3}{4}\left[\text{but area cannot be negative}\right]$
Hence, the required area of $\Delta$ ADE is $\frac{3}{4}$ sq.units.