Question

Let $A(a,0),B(b,2b+1)$ and $C(0,b),b\ne 0,|b|\ne 1,$ be points such that the area of triangle $ABC$ is $1$ sq. unit, then the sum of all possible values of $a$ is

• A. $\frac{2b}{b+1}$
• B. $\frac{-2b^2}{b+1}$
• C. $\frac{2b^2}{b+1}$
• D. $\frac{-2b}{b+1}$

Answer 1

The correct option is B $\frac{-2b^2}{b+1}$

Area $=\left|\begin{array}{c}\frac{1}{2}\left|\begin{array}{ccc}a& 0& 1\\ b& 2b+1& 1\\ o& b& 1\end{array}\right|\end{array}\right|=1$
$\Rightarrow \left(a\right(2b+1-b)-b(-b\left)\right)=\pm 2$
$\Rightarrow a(b+1)=\pm 2-b^2$
$\Rightarrow a=\frac{2-b^2}{b+1}\text{or}\frac{-2-b^2}{b+1}$

Sum of values of $a=\frac{-2-b^2+2-b^2}{b+1}=\frac{-2b^2}{b+1}$