Let two points be A(1,-1) and B(0,2).

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:

1. (1) 4

2. (2) 3

3. (3) 1

4. (4) $-3$

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 $3 x^{\prime}+y^{\prime}=-8$

$\therefore \quad \lambda=3,-2$