Question:
Let two points be $A(1,-1)$ and $B(0,2)$. If $a$ point $P\left(x^{\prime}, y^{\prime}\right)$ be such that the area of $\triangle P A B=5 \mathrm{sq}$. units and it lies on the line, $3 x+y-4 \lambda=0$, then a value of $\lambda$ is:
Correct Option: , 2
Solution:
$D=\frac{1}{2}\left|\begin{array}{rrr}0 & 2 & 1 \\ 1 & -1 & 1 \\ x^{\prime} & y^{\prime} & 1\end{array}\right|=5$
$\Rightarrow-2\left(1-x^{\prime}\right)+\left(y^{\prime}+x^{\prime}\right)=\pm 10$
$\Rightarrow-2+2 x^{\prime}+y^{\prime}+x^{\prime}=\pm 10$
$\Rightarrow 3 x^{\prime}+y^{\prime}=12$ or $3 x^{\prime}+y^{\prime}=-8$
$\therefore \quad \lambda=3,-2$
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