Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2)$. If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\Delta PAB=5$ sq. units and it lies on the line, $3 x+y-4 \lambda=0$, then a value of $\lambda$ is:

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Straight Lines

Let two points be

A (1, - 1)

and

B (0, 2)

. If a point

P (x^{'}, y^{'})

be such that the area of

Δ P A B = 5

sq. units and it lies on the line,

3 x + y - 4 λ = 0

, then a value of

λ

is:

(A)3✓ Correct
(B)4
(C)1
(D)-3

Answer: (A)

Solution

The triangle with vertices

A (1, - 1)

B (0, 2)

and

P (x^{'}, y^{'})

has area

Δ = \frac{1}{2} x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})

. Substituting gives

Δ = \frac{1}{2} 1 (2 - y^{'}) + 0 + x^{'} (- 1 - 2) = \frac{1}{2} 2 - y^{'} - 3 x^{'}

, and setting this to

5

gives

∣3 x^{'} + y^{'} - 2∣ = 10

, so

3 x^{'} + y^{'} = - 8

3 x^{'} + y^{'} = 12

. Since

P

lies on

3 x + y - 4 λ = 0

3 x^{'} + y^{'} = 4 λ

, hence

4 λ = - 8

4 λ = 12

, giving

λ = - 2

λ = 3

; the admissible value is

λ = 3

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option

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