The distance between the point $(1,2)$ from the line $3 x+4 y=6$ measured parallel to the line $3 x-4 y=5$.

• A

  $\sqrt{5}$
• B

  $\frac{\sqrt{5}}{3}$
• C

  $\frac{3}{5}$
• Correct answer
  D

  $\frac{5}{3}$

Solution

Perpendicular distance of point $(1,2)$ to the line $3 x+4 y=6$ is
$=\left|\frac{3 \times 1+4 \times 2-6}{\sqrt{9+16}}\right|=\frac{5}{5}=1$
$\therefore$ Slope of line $3 x-4 y=5$ is $\frac{3}{4}$.
$\begin{aligned}<br/>& \Rightarrow \quad \tan \theta=\frac{3}{4} \\<br/>& \therefore \quad \sin \theta=\frac{3}{5}<br/>\end{aligned}$

From above figure $\sin \theta=\frac{1}{d}$
$\begin{array}{ll}<br/>\Rightarrow & \frac{3}{5}=\frac{1}{d} \\<br/>\Rightarrow & d=\frac{5}{3}<br/>\end{array}$

Hence, the correct answer is option: 'D'.