Question

If $(-4,0)$ and $(1,-1)$ are two vertices of a triangle whose area is $4\text{sq.units}$, then locus of centroid of the triangle is

• A. $3x+15y+4=0$
• B. $x+5y=4$
• C. $3x+15y+20=0$
• D. $x+5y+12=0$

Answer 1

Answer: (A), (C)

$3x+15y+20=0$

Let $(a,b)$ is the third vertex
$\therefore \frac{1}{2}\left|\begin{array}{cccc}-4& 1& a& -4\\ 0& -1& b& 0\end{array}\right|=4$
$\Rightarrow |4+a+b+4b|=8$
$\Rightarrow a+5b+4=\pm 8\Rightarrow a+5b=4\text{or}a+5b=-12\cdots \left(A\right)$
Now,
Centroid$(x,y)\equiv (\frac{a-4+1}{3},\frac{b-1+0}{3})$
$\Rightarrow (x,y)\equiv (\frac{a-3}{3},\frac{b-1}{3})$
$\Rightarrow (a,b)\equiv (3x+3,3y+1)$
putting values of $a$ and $b$ in equation $A$
$\therefore (3x+3)+5(3y+1)=4\Rightarrow 3x+15y+4=0$
or,
$(3x+3)+5(3y+1)+12=0\Rightarrow 3x+15y+20=0$