Question

The distance between the lines $y=2x+4$ and $3y=6x-5$ is equal to

• A. $1$
• B. $3/\sqrt{5}$
• C. $\frac{17\sqrt{5}}{15}$
• D. $\frac{17}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{17\sqrt{5}}{15}$

Given lines are $y=2x+4$ and $3y=6x-5$

Which can be written as $y=2x+4$ and $y-2x+\frac{5}{3}=0$

Thus, the lines $y-2x-4=0$ and $y-2x+\frac{5}{3}=0$ are parallel.
Thus, the distance $=\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|=\left|\frac{\frac{5}{3}+4}{\sqrt{4+1}}\right|=\frac{17\sqrt{5}}{15}$