Explanation
The given points are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,1 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 12,2 \right) \& C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,2 \right)$$
Let us obtain $$ar\Delta ABC$$ by applying the area formula.
$$ ar\Delta =\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right\}$$
$$\Rightarrow ar\Delta =\dfrac { 1 }{ 2 } \left\{ 3\left( 2-2 \right) +12\left( 2-1 \right) +0\left( 1-2 \right) \right\}$$ units $$=6$$ units
$$\therefore ar\Delta =6$$ units
Hence, option B.
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