Question

The distance between the lines $3x+4y=9$ , $6x+8y=15$ is

• A. $\frac{3}{10}$
• B. $\frac{7}{10}$
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

Given, lines are $3x+4y=9$
and $6x+8y=15$ 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
$=\frac{|c_1-c_2|}{\sqrt{a_1^2+b_1^2}}=\frac{|9-\frac{15}{2}|}{\sqrt{3^2+4^2}}$
$=\frac{|18-15|}{2\sqrt{9+16}}$
$=\frac{3}{2\sqrt{25}}=\frac{3}{2\times 5}$
$=\frac{3}{10}$