Question

Distance between the lines $5x+3y-7=0$ and $15x+9y+14=0$ is:

• A. $\frac{35}{\sqrt{34}}$
• B. $\frac{1}{3\sqrt{34}}$
• C. $\frac{35}{3\sqrt{34}}$
• D. $\frac{35}{2\sqrt{34}}$

Answer 1

Answer: (C)

Given lines are $5x+3y-7=0$
and $15x+9y+14=0$
The distance from origin to the lines are
$d_1=\frac{0+7-7}{\sqrt{5^2+3^2}}=\frac{-7}{\sqrt{34}}$
and $d_2=\frac{0+0+14}{\sqrt{225+81}}$
$=\frac{14}{\sqrt{306}}=\frac{14}{3\sqrt{34}}$
Since, the distance is on opposite sign, it means that the given lines are on opposite side of the origin, therefore the distance between them is
$d_1+d_2=\frac{7}{\sqrt{34}}+\frac{14}{3\sqrt{34}}=\frac{35}{3\sqrt{34}}$
Alternative Solution:
Given lines are $5x+3y-7=0$
and $15x+9y+14=0$ or $5x+3y+\frac{14}{3}=0$
Since, the given lines are parallel, therefore the distance between them is
$d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{|-7-\frac{14}{3}|}{\sqrt{5^2+3^2}}$
$=\frac{35}{3\sqrt{34}}$
Let $ax+by+c_1=0$ and $ax+by+c_2=0$ are two parallel lines,
then the distance between them is
$d=\frac{c_1-c_2}{\sqrt{a^2+b^2}}$