Question

Let $3x+4y-12=0$ be a given line and origin be shifted at $(4,5)$ then the distance between the old and new line be

• A. $\frac{8}{5}$
• B. $\frac{32}{25}$
• C. $\frac{32}{5}$
• D. $\frac{12}{5}$

Answer 1

The correct option is C $\frac{32}{5}$

Equation of old line is $3x+4y-12=0$ as origin is shifted at $(4,5)$
$\therefore$ Equation of new line be $3(x+4)+4(y+5)-12=0$
ie new line is $3x+4y+20=0$
Now new line and old lines are parallel and distance between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by
$\therefore d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{20+12}{\sqrt{3^2+4^2}}=\frac{32}{5}$