Question

The equation of one of the straight lines passing through the point $(0,1)$ and is at a distance of $\frac{3}{5}$ units from the origin is

• A. $4x+3y=3$
• B. $-x+y=1$
• C. $x+y=1$
• D. $5x+4y=4$
• E. $-5x+4y=4$

Answer 1

The correct option is A $4x+3y=3$

The equation is $(y-1)=m(x-0)$
$\Rightarrow \frac{1}{\sqrt{m^2+1}}=\frac{3}{5}[\because \left|\frac{x_{1}a+y_{1}b+}{\sqrt{a^2+b^2}}\right|]$
$\Rightarrow m^2+1=\frac{25}{9}\Rightarrow m=\pm \frac{4}{3}$
Thus, the required equation of line is
Then, $-\frac{4}{3}x-y+1=0$ (From Eq.(i)]
$4x+3y-3=0$
$4x+3y=3$