Question

Question 16
If D $(\frac{-1}{2},\frac{5}{2})$, E (7,3) and F $(\frac{7}{2},\frac{7}{2})$ are the mid-points of sides of $\Delta ABC$, then find the area of the $\Delta ABC$.

Answer 1

$LetA=(x_1,y_1),B=(x_2,y_2)andC=(x_3,y_3)\text{are the vertices of the}\Delta ABC$

Given:
$D(-\frac{1}{2},\frac{5}{2})\text{is the mid-point of BC.}$
$\therefore \frac{x_2+x_3}{2}=-\frac{1}{2}and\frac{y_1+y_2}{2}=\frac{5}{2}\left[\begin{array}{c}\because \text{Mid-point of a line segment having points}\\ (x_1,y_1)and(x_2,y_2),\\ =(\frac{x_1+x_2}{2},\frac{y_2+y_3}{2})\end{array}\right]\Rightarrow x_2+x_3=-1...\text{(i)}And{y}_2+y_3=5\text{(ii)}\text{As E(7,3) is the mid-point of CA.}\therefore \frac{x_3+x_1}{2}=7And\frac{y_3+y_1}{2}=3\Rightarrow x_3+x_1=14...\text{(iii)}And{y}_3+y_1=6...\text{(iv)}Also,F(\frac{7}{2},\frac{7}{2})\text{is the mid-point of AB.}\therefore \frac{x_1+x_2}{2}=\frac{7}{2}And\frac{y_1+y_2}{2}=\frac{7}{2}\Rightarrow x_1+x_2=7...\text{(v)}And{y}_1+y_2=7...\text{(vi)}$

$\text{On adding Eq.(i) (iii) and (v), we get}2(x_1+x_2+x_3)=20$
$\Rightarrow x_1+x_2+x_3=10...\text{(vii)}$

$\text{On substracting Eqs.(i), (iii) and (v) from Eq. (vii)}\text{respectively, we get}{x}_1=1,x_2=-4,x_3=3\text{On adding Eq.(ii) (iv) and (vi),we get}2(y_1+y_2+y_3)=18$
$\Rightarrow y_1+y_2+y_3=9$. . . .(viii)
$\text{On substracting Eqs.(ii), (iv) and (vi) from Eq. (viii)}\text{respectively,we get}{y}_1=4,y_2=3,y_3=2$

$\text{Hence, the vertices of}\Delta ABC\text{are}$ $A(11,4),B(-4,3)andC(3,2)$

$\because Areaof\Delta ABC=\Delta =\frac{1}{2}\left[x\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\therefore \Delta =\frac{1}{2}\left[11\right(3-2)+(-4\left)\right(2-4)+3(4-3\left)\right]=\frac{1}{2}[11\times 1+(-4\left)\right(-2)+3(1\left)\right]=\frac{1}{2}[11+3+8]=\frac{22}{2}=11\therefore Requiredareaof\Delta ABC=11$