Question

Let medians $L_1,L_2$ and $L_3$ of a $△ABC$ belong to the family of lines $2x-y+2+\lambda (x+2y+1)=0$ where $\lambda$ a parameter. Points $P,Q,R$ are the mid-points of sides BC, CA and AB respectively where $P\equiv (2,6)$ are $Q\equiv (-5,2)$. If area of $△BRG$ is $6\lambda$ where $G$ is centroid of $△ABC$, then find the value of $\lambda$

Answer 1

Let $R$ and $B$ have coordinates $(h,k)$ and $(p,q)$ respectively.

Since the family of lines represents the medians of $△ABC$, their intersection point is the centroid($G$) of $△ABC$.

$2(2x-y+2)+x+2y+1=0⟹x=-1,y=0⟹G=(-1,0)$

Now, the centroid of $△ABC$ is also the centroid of its pedal $△PQR$.

$\therefore (-1,0)=(\frac{h-5+2}{3},\frac{2+6+k}{3})⟹(h,k)=(0,-8)$

Also, as centroid divides median in $2:1$ ratio, we have

$\frac{BG}{GQ}=2⟹(-1,0)=(\frac{2\times -5+p\times 1}{3},\frac{2\times 2+q}{3})$

$⟹(p,q)=(7,-4)$

$\therefore$ area of $△BRG=30$ (using heron's formula)