# The line 4x+3y-12equal to 0 intersects x y at A B respectively

Please find below the solution to the asked query :

$\mathrm{Area}\mathrm{of}\mathrm{triangle}\mathrm{AOB}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 3\times 4\phantom{\rule{0ex}{0ex}}=6\mathrm{sq}.\mathrm{units}$

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