Q41 / 160 · AP EAMCET 2014 · Mathematics
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
- A$x-2 y=6$
- Correct answerB$4 x+3 y=12$
- C$4 x+3 y+24=0$
- D$3 x+4 y=12$
Solution
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 $$ 4 x+3 y= \pm 12 $$
$\therefore$ 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 $$ 4 x+3 y= \pm 12 $$
Hence, the correct answer is option: 'B'.