Let two points be A(1, - 1) and B(0, 2) . If a point P(x', 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

Correct answer: B. 3 ∆ = (1)/(2)| cases 0 & 2 & 1 1 & - 1 & 1 x' & y' & 1 cases | ⇒ - 2(1 - x') + (y' + x') = ± 10 ⇒ - 2 + 2x' + y' + x' = ± 10 ⇒ 3x' + y' = 12 Or 3x' + y' = - 8 . . . (i) (x', y') lies on the line 3x + y - 4λ = 0 , So, 3x' + y' - 4λ = 0 . . . (ii) Solving (i) and (ii) , we get, λ = 3, - 2

JEE Main 20208 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2020 · 8 January · Morning Shift · Mathematics · Straight Lines

Let two points be A(1, - 1) and B(0, 2) . If a point P(x', y') be...

Let two points be

A\left(1, - 1\right)

and

B\left(0, 2\right) .

If a point

P\left(x\text{'}, y\text{'}\right)

be such that the area of

∆ PAB = 5

sq. units and it lies on the line

3x + y - 4\lambda = 0,

then a value of

\lambda

Options

A $4$
B $3$
C $1$
D $- 3$

Correct answer

B. $3$

Step-by-step solution

∆ = \dfrac{1}{2}\left| $\begin{cases}0 & 2 & 1 \\ 1 & - 1 & 1 \\ x\text{'} & y\text{'} & 1\end{cases}$ \right| $ $ ⇒ - 2\left(1 - x\text{'}\right) + \left(y\text{'} + x\text{'}\right) = \pm 10 $ $ ⇒ - 2 + 2x\text{'} + y\text{'} + x\text{'} = \pm 10 $ $ ⇒ 3x\text{'} + y\text{'} = 12 $Or$ 3x\text{'} + y\text{'} = - 8 . . . \left(i\right) $\left(x\text{'}, y\text{'}\right)$ lies on the line $3x + y - 4\lambda = 0$ ,

So, $3x\text{'} + y\text{'} - 4\lambda = 0 . . . \left(ii\right)$

Solving \left(i\right) $and \left(ii\right)$ , we get,
$\lambda = 3, - 2$

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