Let two points be A 1 , - 1 and B 0 , 2 . If a point P x ' ,   y ' be such that the area of ∆ P A…
Let two points be and If a point be such that the area of sq. units and it lies on the line then a value of is
Solution
$\Delta = \frac{1}{2} \begin{vmatrix} 0 & 2 & 1 \\ 1 & -1 & 1 \\ x' & y' & 1 \end{vmatrix}$
$\Rightarrow -2 \begin{vmatrix} 1 & -x' \end{vmatrix} + \begin{vmatrix} y' & x' \end{vmatrix} = \pm 10$
$\Rightarrow -2 + 2x' + y' + x' = \pm 10$
$\Rightarrow 3x' + y' = 12$ or $3x' + y' = -8$
$(x', y')$ lies on the line $3x + y - 4\lambda = 0$,
So, $3x' + y' - 4\lambda = 0$.
Solving $i$ and $ii$, we get,
$\lambda = 3, -2$
Asked in: JEE Main 2020 (08 Jan Shift 1)
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