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- 3 x - y -4 -0 - then the value of - is-A- 4B- 1C- -3D- 3

Area of triangle is A = 12 1 amp; 1 amp; 1 0 amp; 2 amp; 1 x' amp; y' amp; 1 = ± 5 ⇒ (2 y') x' 2x' = ± 10 ⇒ 3x' y' + 2 = ± 10 ⇒ 3x' ...

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Byju's Answer

Standard XII

Mathematics

Area of Triangle with Coordinates of Vertices Given

Let two point...

Question

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 the value of $λ$ is:

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Solution

The correct option is D $3$
Area of triangle is
$A = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & 1 0 & 2 & 1 x^{'} & y^{'} & 1 \end{matrix} ∣ ∣ ∣ ∣ = \pm 5$

$\Rightarrow (2 - y^{'}) - x^{'} - 2 x^{'} = \pm 10$
$\Rightarrow - 3 x^{'} - y^{'} + 2 = \pm 10$
$\Rightarrow 3 x^{'} + y^{'} = 12$ or $3 x^{'} + y^{'} = - 8$
$\Rightarrow λ = 3$ or $λ = - 2$

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Similar questions

Q. 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 the value of

λ

is:

Q. The point

P (x, y)

lies on the line

2 x + 3 y + 1 = 0

such that

| P A - P B |

is maximum where

A (2, 0), B (0, 2)

Q. The vertices

B

and

C

of a

△ A B C

lie on the line,

\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}

such that

B C = 5

units. Then the area (in sq. units) of this triangle, given that the point

A (1, - 1, 2)

, is :

Q. Let

O

be the origin, and let

A (1, 0), B (0, 1)

be two points. If

P (x, y)

is a point such that

x y > 0

and

x + y < 1,

then

Q. The vertices

B

and

C

of a

△ A B C

lie on the line,

\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}

such that

B C = 5

units. Then the area (in sq. units) of this triangle, given that the point

A (1, - 1, 2)

is:

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Let two points be A(1,−1) and B(0,2). If a point P(x′,y′) be such that the area of ΔPAB=5 sq. units and it lies on the line, 3x+y−4λ=0, then the value of λ is:

The correct option is D 3Area of triangle is A=12∣∣ ∣∣1−11021x′y′1∣∣ ∣∣=±5 ⇒(2−y′)−x′−2x′=±10 ⇒−3x′−y′+2=±10 ⇒3x′+y′=12 or 3x′+y′=−8 ⇒λ=3 or λ=−2

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 the value of $λ$ is: