If O is the origin and Px1,y1,Qx2,y2 are two points OP×OQsin⁡∠POQ=

We, have,Area of ΔOPQ=Absolute value of 120 0 1x1 y1 1x2 y2 1⇒ Area of ΔOPQ=12x1y2−x2y1Also, Area of ΔOPQ=12OP×OQ×sin⁡∠POQ∴ 12OP×OQsin⁡∠POQ=12x1y2−x2y1⇒ OP×OQsin⁡∠POQ=x1y2−x2y1

If O is the origin and Px1,y1,Qx2,y2 are two points OP×OQsin⁡∠POQ=

We, have,Area of ΔOPQ=Absolute value of 120 0 1x1 y1 1x2 y2 1⇒ Area of ΔOPQ=12x1y2−x2y1Also, Area of ΔOPQ=12OP×OQ×sin⁡∠POQ∴ 12OP×OQsin⁡∠POQ=12x1y2−x2y1⇒ OP×OQsin⁡∠POQ=x1y2−x2y1

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If O is the origin and $P (x_{1}, y_{1}), Q (x_{2}, y_{2})$ are two points $O P \times O Q \sin ∠ P O Q =$

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none of these

$x_{1} y_{2} + x_{2} y_{1}$

$|x_{1} y_{2} - x_{2} y_{1}|$

$x_{1} x_{2} + y_{1} y_{2}$

answer is C.

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Detailed Solution

We, have,

Area of $ΔO P Q =$ Absolute value of $\frac{1}{2} |\begin{array}{lll} 0 0 1 \\ x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \end{array}|$

$\Rightarrow$ Area of $Δ O P Q = \frac{1}{2} |x_{1} y_{2} - x_{2} y_{1}|$

Also, Area of $Δ O P Q = \frac{1}{2} O P \times O Q \times \sin ∠ P O Q$

$\begin{array}{ll} ∴ \frac{1}{2} O P \times O Q \sin ∠ P O Q = \frac{1}{2} |x_{1} y_{2} - x_{2} y_{1}| \\ \Rightarrow O P \times O Q \sin ∠ P O Q = |x_{1} y_{2} - x_{2} y_{1}| \end{array}$