Solve this


The area of $\triangle A B O$ with vertices $A(a, 0), O(0,0)$ and $B(0, b)$ in square units is

(a) $a b$

(b) $\frac{1}{2} a b$

(c) $\frac{1}{2} a^{2} b^{2}$

(d) $\frac{1}{2} b^{2}$


Let A(x1 = ay1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So

Area $(\Delta A B O)=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$


$=\frac{1}{2}|-a b|=$

$=\frac{1}{2} a b$

Hence, the correct answer is (b).

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