Cartesian plane

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Let ABC be a triangle whose vertices, $A (1, - 1), B (0, 2) C (x^{'}, y^{'})$ such that area of $△ A B C is 5 sq .$ and $C (x^{'}, y^{'}) lie on 3 x + y - 4 λ = 0 .$ $Then, λ =$

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It is given that area of $△ A B C is 5$ sq. units
$\begin{array}{l} ∴ \frac{1}{2} |\begin{array}{ccc} 1 & - 1 & 1 \\ 0 & 2 & 1 \\ x^{'} & y^{'} & 1 \end{array}| = \pm 5 \\ \Rightarrow (2 - y^{'}) + (0 - x^{'}) + (0 - 2 x^{'}) = \pm 10 \\ \Rightarrow - 3 x^{'} - y^{'} + 2 = \pm 10 \\ \Rightarrow 3 x^{'} + y^{'} - 2 = \pm 10 \\ \Rightarrow 3 x^{'} + y^{'} - 12 = 0 or 3 x^{'} + y^{'} + 8 = 0 \end{array}$
Point $C (x^{'}, y^{'}) lies on 3 x + y - 4 λ = 0 . Therefore, λ = 3$

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Cartesian plane

Remember concepts with our Masterclasses.

Let ABC be a triangle whose vertices, A(1,−1),B(0,2)Cx′,y′ such that area of △ABC is 5sq. and Cx′,y′ lie on 3x+y−4λ=0. Then, λ=

It is given that area of △ABC is 5 sq. units∴ 121−11021x′y′1=±5⇒ 2−y′+0−x′+0−2x′=±10⇒ −3x′−y′+2=±10⇒ 3x′+y′−2=±10⇒ 3x′+y′−12=0 or 3x′+y′+8=0Point Cx′,y′ lies on 3x+y−4λ=0. Therefore, λ=3

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Let ABC be a triangle whose vertices, $A (1, - 1), B (0, 2) C (x^{'}, y^{'})$ such that area of $△ A B C is 5 sq .$ and $C (x^{'}, y^{'}) lie on 3 x + y - 4 λ = 0 .$ $Then, λ =$