Write down the equation of the line AB, through (3,2), perpendicular to the line 2y= 3x+5. AB meets the x-axis at A and y-axis at B. Write down the coordinates of A and B. Calculate the area of triangle OAB where O is the origin.  

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Please find below the solution to the asked query:
$$\begin{gathered} \Rightarrow 2\mathrm{y}=3\mathrm{x}+5 \\ \Rightarrow \mathrm{y}=\frac{3}{2}\mathrm{x}+\frac{5}{2} \\ \mathrm{Compare} \mathrm{with} \mathrm{y}=\mathrm{mx}+\mathrm{c} \\ \Rightarrow \mathrm{m}1\left(\mathrm{slope}\right)=\frac{3}{2} \\ \mathrm{Now} \mathrm{perpendicular} \mathrm{line} \mathrm{have} \mathrm{slope} \mathrm{m}1\times \mathrm{m}2=-1 \\ \mathrm{So} \mathrm{m}2=\frac{-2}{3} \\ \mathrm{So} \mathrm{line} \mathrm{passes} \mathrm{through}\left(3,2\right) \\ \Rightarrow \mathrm{y}-\mathrm{y}1=\mathrm{m}2\left(\mathrm{x}-\mathrm{x}1\right) \\ \Rightarrow \mathrm{y}-2=\frac{-2}{3}\left(\mathrm{x}-3\right) \\ \Rightarrow 3\mathrm{y}-6=-2\mathrm{x}+6 \\ \Rightarrow 2\mathrm{x}+3\mathrm{y}=12...\left(1\right) \mathrm{AB} \mathrm{line} \\ \mathrm{At} \mathrm{x}-\mathrm{axis} \mathrm{y}=0 \\ \Rightarrow 2\mathrm{x}=12 \\ \Rightarrow \mathrm{x}=6 \\ \mathrm{So} \mathrm{point} \mathrm{A} \mathrm{is} \left(6,0\right) \\ \mathrm{At} \mathrm{y}-\mathrm{axis} \mathrm{x}=0 \\ \Rightarrow 3\mathrm{y}=12 \\ \Rightarrow \mathrm{y}=4 \\ \mathrm{So} \mathrm{point} \mathrm{B} \mathrm{is} \left(0,4\right) \\ \mathrm{Now} \mathrm{Area} \mathrm{of} \mathrm{OAB}, \left(0,0\right),\left(0,4\right),\left(6,0\right) \\ \mathrm{So} \mathrm{length} \mathrm{OA}=6 \mathrm{unit} \mathrm{and} \mathrm{OB}=4\mathrm{unit} \\ \mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 6\times 4=12 {\mathrm{unit}}^2 \mathrm{Answer} \end{gathered}$$

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