If origin is centre of gravity of a triangle whose two vertices are are cos- sin- and cos- sin- then maximum area of such triangle is

Centroid divides area of ABC into three equal parts ∴ GBC=13 ABC=12|x_1y_2 x_2y_1| Here (x_1,y_1)=(cosα, nbsp; sinα) and (x_2,y_2)=(cosβ, sinβ) ∴ ABC=32sin (α ...

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Standard XII

Mathematics

Relation between Slopes and Coefficients in Equations of Pair of Lines

If origin is ...

Question

If origin is centre of gravity of a triangle whose two vertices are are $(cos α, sin α)$ and $(cos β, sin β)$ then maximum area of such triangle is

1

sq. units

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1.5

sq. units

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2

sq. units

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2.5

sq. units

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Solution

The correct option is B $1.5$ sq. units
Centroid divides area of $△ A B C$ into three equal parts
$∴ △ G B C = \frac{1}{3} △ A B C = \frac{1}{2} | x_{1} y_{2} - x_{2} y_{1} |$
Here $(x_{1}, y_{1}) = (cos α, sin α)$ and $(x_{2}, y_{2}) = (cos β, sin β)$
$∴ △ A B C = \frac{3}{2} sin (α - β)$
$△ A B C_{m a x} = \frac{3}{2} = 1.5$ sq. units

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If origin is centre of gravity of a triangle whose two vertices are are (cosα,sinα) and (cosβ,sinβ) then maximum area of such triangle is

The correct option is B 1.5 sq. unitsCentroid divides area of △ABC into three equal parts ∴△GBC=13△ABC=12|x1y2−x2y1| Here (x1,y1)=(cosα,sinα) and (x2,y2)=(cosβ,sinβ) ∴△ABC=32sin(α−β) △ABCmax=32=1.5 sq. units

If origin is centre of gravity of a triangle whose two vertices are are $(cos α, sin α)$ and $(cos β, sin β)$ then maximum area of such triangle is