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

Answer: (A)

The equation is $(y-1)=m(x-0)$
$\Rightarrow mx-y+1=0$
$\Rightarrow \frac{1}{\sqrt{m^2+1}}=\frac{3}{5}$
$[\because d=\left|\frac{x_{1}a+y_{1}b+c}{\sqrt{a^2+b^2}}\right|$ at point $\left(x_1,y_1\right)]$
$\Rightarrow m^2+1=\frac{25}{9}\Rightarrow m^2=\frac{16}{9}\Rightarrow m=\pm \frac{4}{3}$
Thus, the required equation of line is Then,
$-\frac{4}{3}x-y+1=0$ [from Eq.(i) ]
$\Rightarrow 4x+3y-3=0\Rightarrow 4x+3y=3$