The equation of a straight line, perpendicular to $3 x-4 y=6$ and forming a triangle of area 6 sq. units with coordinate axes, is

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Aicharya · Accepted Answer

Given equation of line is $ 3 x-4 y=6 $ Perpendicular to above line is $ 4 x+3 y=k $

$\therefore$ Area of $\triangle O A B$ $ =\frac{1}{2} \times O A \times O B $ $ \begin{aligned} \Rightarrow & & 6 & =\frac{1}{2} \times \frac{k}{4} \times \frac{k}{3} \\ & \Rightarrow & 6 & =\frac{k^2}{24} \\ & \Rightarrow & k^2 & =144 \\ & \Rightarrow & k & = \pm 12 \end{aligned} $ $\therefore$ Required equation of line is $ 4 x+3 y= \pm 12 $

The equation of a straight line, perpendicular to $3 x-4 y=6$ and forming a triangle of area 6 sq. units…

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