Q65 / 200 · TS EAMCET 2002 · Mathematics
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 :
- A$3 x+2 y= \pm 2$
- Correct answerB$3 x+2 y= \pm 6$
- C$3 x+2 y= \pm 4$
- D$3 x+2 y= \pm 8$
Solution
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) $$ 3 x+2 y= \pm 6 $$
Hence, the correct answer is option: 'B'.