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Q.
Distance between the lines $ 5x+3y-7=0 $ and $ 15x+9y+14=0 $ is:
Jharkhand CECEJharkhand CECE 2002
Solution:
Given lines are $5 x+3 y-7=0$
and $15 x+9 y+14=0$
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 $5 x+3 y-7=0$
and $15 x+9 y+14=0$ or $5 x+3 y+\frac{14}{3}=0$
Since, the given lines are parallel, therefore the distance between them is
$d=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{\left|-7-\frac{14}{3}\right|}{\sqrt{5^{2}+3^{2}}} $
$=\frac{35}{3 \sqrt{34}}$
Let $a x+b y+c_{1}=0$ and $a x+b y+c_{2}=0$ are two parallel lines,
then the distance between them is
$d=\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}$