If a straight line perpendicular to $2 x-3 y+7=0$ forms a triangle with the co-ordinate axes whose area is 3 sq. units, then the equation of the straight line is :

Question

Aicharya · Accepted Answer

The given line is $2 x-3 y+7=0$

A perpendicular line to given line is $ 3 x+2 y+k=0 $ Put $y=0 \Rightarrow 3 x+k=0$ $ x=-\frac{k}{3} $ Put $x=0 \Rightarrow 0+2 y+k=0$ $ y=-\frac{k}{2} $ Area of the triangle $O A B=\frac{1}{2} \cdot O A \cdot O B$ $ \begin{aligned} \Rightarrow & 3 & =\frac{1}{2} \cdot\left(-\frac{k}{3}\right)\left(-\frac{k}{2}\right) \\ \Rightarrow & k^2 & =36 \Rightarrow k= \pm 6 \end{aligned} $ Putting in (i) $ 3 x+2 y= \pm 6 $

If a straight line perpendicular to $2 x-3 y+7=0$ forms a triangle with the co-ordinate axes whose area is 3…

Solution