If a straight line $L$ perpendicular to the line $3 x-4 y=6$ forms a triangle of area 6 square units with…
If a straight line $L$ perpendicular to the line $3 x-4 y=6$ forms a triangle of area 6 square units with coordinate axes, then the minimum perpendicular distance from the point $(1,1)$ to the line $\mathrm{L}$ is
1
$\sqrt{2}$
2
$\sqrt{3}$
Solution
The line which is perpendicular to the given line will have a slope of $-\frac{4}{3}$.
The equation of the line will look like $3 y+4 x=k$.
The line intersects the $\mathrm{x}$-axis at $\left(\frac{\mathrm{k}}{4}, 0\right)$ an $\mathrm{y}$-axis at $\left(0, \frac{\mathrm{k}}{3}\right)$.
The area of the triangle formed is given by
$
\begin{aligned}
& =\frac{1}{2} \times \frac{k}{3} \times \frac{k}{4} \Rightarrow 6=\frac{1}{2} \times \frac{k^2}{12} \\
& \Rightarrow k^2=12 \times 12 \Rightarrow k= \pm 12
\end{aligned}
$
Thus, the equation of line becomes $4 x+3 y= \pm 12$ or $4 x+3 y \pm 12=0$
For minimum distance from point $(1,1)$, we will take negative sign.
So, the equation of line is $4 x+3 y-12=0$
The required distance $=\left|\frac{4 \times 1+3 \times 1-12}{\sqrt{4^2+3^2}}\right|$
$
=\left|\frac{4+3-12}{5}\right|=\frac{5}{5} \Rightarrow=1
$