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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 :
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Verified Answer
The correct answer is:
$3 x+2 y= \pm 6$
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)
$$
3 x+2 y= \pm 6
$$
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)
$$
3 x+2 y= \pm 6
$$
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