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 A1,-1 and B0,2. If a point Px', y' be such that the area of PAB=5 sq. units and it lies on the line 3x+y-4λ=0, then a value of λ is
  1. 4
  2. 3
  3. 1
  4. -3

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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