Question

In a triangle $ABC$ if the equation of the median $AD$ and $BE$ are $2x+3y-6=0$ and $3x-2y-10=0$ respectively and $AD=6,BE=11$, then find the area of the triangle $ABC$.

Answer 1

Ref. Image.

Let AD & BE

intersect as cx

slope of AD is

$m_1=\frac{-2}{3}$ as its equation is $2x+3y-5=0$

slope of BC is

$m_2=\frac{3}{2}$ as equation is $3x-2y-10=0$

As $m_1\times m_2=-1$, so angle between medians

AD & BE at G is $\angle$ AGB =${90}^{\circ }$

Now ,

$AG=\frac{2}{3}\times$ AD $=\frac{2}{3}\times 6=4$

$BG=\frac{2}{3}\times$ BE $=\frac{2}{3}\times 11=\frac{22}{3}$

area of $\Delta$ ABG $=\frac{1}{2}\times AG\times BG$

$=\frac{1}{2}\times 4\times \frac{22}{3}$

$=\frac{44}{3}$ sq . units .

$\therefore$ area of $\Delta$ ABC = $3\times \frac{44}{3}=44$ sq. units.