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