Question

The vertices of a triangle ABC are $(\lambda ,2-2\lambda ),(-\lambda +1,2\lambda )$ and $(-4-\lambda ,6-,2\lambda )$. If its area be 10 units then number of integral values of $\lambda$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Area $=\frac{1}{2}\left|\begin{array}{ccc}\lambda & 2-2\lambda & 1\\ -\lambda +1& 2\lambda & 1\\ -4-\lambda & 6-2\lambda & 1\end{array}\right|=70$
$=\frac{1}{2}\left|2{\lambda }^2+(6-2\lambda )(1-\lambda )-(4+\lambda )(2-2\lambda )-(-2\lambda )(\lambda +4)+\lambda (6-2\lambda )-(2-2\lambda )(1-\lambda )\right|=70$
$=\mid 2{\lambda }^2+6-2\lambda -6\lambda +2{\lambda }^2-8-2\lambda +8\lambda +2{\lambda }^2+2{\lambda }^2-8\lambda +6\lambda -2{\lambda }^2+2-2\lambda -2\lambda -2\lambda =140$
$=\left|8{\lambda }^2+6\lambda -4\right|=140$
$\Rightarrow 2{\lambda }^2+8\lambda -9\lambda -36=0$
$\lambda =\frac{9}{2},-4$
$2{\lambda }^2+\lambda -1=-35$
$2{\lambda }^2+\lambda +34=0$
D $<0$ Integral value of $\lambda$
$\therefore \lambda =-4$