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Q.
The distance between the lines $ 3x + 4y = 9 $ , $ 6x + 8y = 15 $ is
AMUAMU 2014Straight Lines
Solution:
Given, lines are $3x + 4y = 9$
and $6 x + 8 y = 15$ or $3x+4y=\frac{15}{2}$
It is clear that given lines are parallel
Here, $C_{1}=9$, $C_{2}=\frac{15}{2}$,
$a_{1}=3$ and
$b_{1}=4$
$\therefore $ Distance between two lines
$=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{\left|9-\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}}$
$=\frac{\left|18-15\right|}{2\sqrt{9+16}}$
$=\frac{3}{2\sqrt{25}}=\frac{3}{2\times5}$
$=\frac{3}{10}$