Question

The minimum distance of the point $(2,-2)$ from the line $y=mx+1$ where $m>0$ is $3$ units. The value of $m$ is

• A. $\frac{11}{5}$
• B. $\frac{12}{5}$
• C. $\frac{9}{5}$
• D. $\frac{5}{12}$

Answer 1

The correct option is B $\frac{12}{5}$

Minimum distance is the perpendicular distance between the point and the line.
Equation of line
$y=mx+1\Rightarrow -mx+y-1=0$
So the perpendicular distance can be written as,
$3=|\frac{-2m-2-1}{\sqrt{1+m^2}}|\Rightarrow \pm 3\left(\sqrt{1+m^2}\right)=-2m-3\Rightarrow 9(1+m^2)=4m^2+9+12m\Rightarrow 5m^2-12m=0\Rightarrow m=0,\frac{12}{5}$
But, as given $m>0$
$\therefore m=\frac{12}{5}$