Question

Which of the following line(s) is nearest to origin

• A. $4x-3y+12=0$
• B. $3x-3y+12=0$
• C. $3x-4y+4=0$
• D. $3x+4y+5=0$

Answer 1

The correct option is C $3x-4y+4=0$

Distance of origin from the line $ax+by+c=0$ will be $\frac{|c|}{\sqrt{a^2+b^2}}$ units
so,
$\left(i\right)$distance of the line $4x-3y+12=0$ from the origin
$=\frac{12}{5}$units

$\left(ii\right)$distance of the line $3x-3y+12=0$ from the origin
$=\frac{12}{3\sqrt{2}}=2\sqrt{2}$units

$\left(iii\right)$distance of the line $3x-4y+4=0$ from the origin
$=\frac{4}{5}$units

$\left(iv\right)$distance of the line $-3x-4y+5=0$ from the origin
$=1$units
$\therefore$ line $3x-4y+4=0$ will be nearest to origin