\(A(6,1),\ B(8,2)\) and \(C(9,4)\) are three vertices of a parallelogram \(ABCD\). If \(E\) is the midpoint of \(DC\), find the area of \(\triangle ADE\).
Area \(= \dfrac{3}{4}\) square units.
Step 1: Let the fourth vertex be \(D(x_4,y_4)\).
Step 2: In a parallelogram, the diagonals bisect each other, so the midpoint of \(BD\) equals the midpoint of \(AC\).
Step 3: Midpoint of \(BD\) is
\(\left(\dfrac{x_4+8}{2},\dfrac{y_4+2}{2}\right)\).
Step 4: Midpoint of \(AC\) is
\(\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{15}{2},\dfrac{5}{2}\right)\).
Step 5: Equate the midpoints:
\(\dfrac{x_4+8}{2}=\dfrac{15}{2}\Rightarrow x_4+8=15\Rightarrow x_4=7\)
\(\dfrac{y_4+2}{2}=\dfrac{5}{2}\Rightarrow y_4+2=5\Rightarrow y_4=3\)
So, \(D=(7,3)\).
Step 6: Since \(E\) is the midpoint of \(DC\),
\(E=\left(\dfrac{7+9}{2},\dfrac{3+4}{2}\right)=\left(\dfrac{16}{2},\dfrac{7}{2}\right)=\left(8,\dfrac{7}{2}\right)\).
Step 7: Now find area of \(\triangle ADE\) using the coordinate area formula:
\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\).
Here \(A(6,1),\ D(7,3),\ E\left(8,\dfrac{7}{2}\right)\):
\(\text{Area} = \dfrac{1}{2}\left|6\left(3-\dfrac{7}{2}\right)+7\left(\dfrac{7}{2}-1\right)+8(1-3)\right|\)
Step 8: Simplify:
\(3-\dfrac{7}{2}=-\dfrac{1}{2},\quad \dfrac{7}{2}-1=\dfrac{5}{2},\quad 1-3=-2\)
\(\Rightarrow \text{Area} = \dfrac{1}{2}\left|6\left(-\dfrac{1}{2}\right)+7\left(\dfrac{5}{2}\right)+8(-2)\right| \)
\( = \dfrac{1}{2}\left|-3+\dfrac{35}{2}-16\right| \)
\(= \dfrac{1}{2}\left|\dfrac{-6+35-32}{2}\right| = \dfrac{1}{2}\left|\dfrac{-3}{2}\right| = \dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Final Answer: \(\text{Area}(\triangle ADE)=\dfrac{3}{4}\) square units.