Question

If $D\left(-\frac{1}{5},\frac{5}{2}\right),E(7,3)\mathrm{and}\mathrm{F}\left(\frac{7}{2},\frac{7}{2}\right)$ are the mid-points of sides of $∆ABC$, find the area of $∆ABC$.

Answer 1

The midpoint of BC is $D\left(-\frac{1}{2},\frac{5}{2}\right)$,
The midpoint of AB is $F\left(\frac{7}{2},\frac{7}{2}\right)$,
The midpoint of AC is $E\left(7,3\right)$,
Consider the line segment BC,
$$\begin{gathered} \Rightarrow \frac{p+r}{2}=-\frac{1}{2};\frac{q+s}{2}=\frac{5}{2} \\ \Rightarrow p+r=-1;q+s=5.....\left(i\right) \\ ConsiderthelinesegmentAB, \\ \Rightarrow \frac{p+x}{2}=\frac{7}{2};\frac{q+y}{2}=\frac{7}{2} \\ \Rightarrow p+x=7;q+y=7.....\left(ii\right) \end{gathered}$$

$$\begin{gathered} ConsiderthelinesegmentAC, \\ \Rightarrow \frac{r+x}{2}=7;\frac{s+y}{2}=3 \\ \Rightarrow r+x=14;s+y=6.....\left(iii\right) \end{gathered}$$

Solve (i), (ii) and (iii) to get $A\left(x,y\right)=A\left(11,4\right),B\left(p,q\right)=B\left(-4,3\right),C\left(r,s\right)=C\left(3,2\right)$
Let us assume that BC is base of the triangle,
$$\begin{gathered} \overline{BC}=\sqrt{{\left(-4-3\right)}^2+{\left(3-2\right)}^2}=\sqrt{50} \\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{BC}\mathrm{is} \\ \frac{\mathrm{x}+4}{-4-3}=\frac{\mathrm{y}-3}{3-2} \\ \Rightarrow \mathrm{x}+7\mathrm{y}-17=0 \\ \mathrm{The}\mathrm{perpendicular}\mathrm{distance}\mathrm{from}\mathrm{a}\mathrm{point}\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\mathrm{is} \\ \mathrm{P}=\left|\frac{1\left(11\right)+7\left(4\right)-17}{\sqrt{50}}\right|=\frac{22}{\sqrt{50}} \end{gathered}$$
The area of the triangle is $A=\frac{1}{2}\times \sqrt{50}\times \frac{22}{\sqrt{50}}=11\mathrm{sq}.\mathrm{units}$