The distance of the point $(3,4)$ from the line $3 x+2 y+7=0$ measured along the line parallel to $y-2 x+7=0$ is equal to

• Correct answer
  A

  $\frac{24 \sqrt{5}}{7}$
• B

  $3 \sqrt{5}$
• C

  $\frac{23 \sqrt{5}}{7}$
• D

  $4 \sqrt{5}$

Solution

The slope of the line, $y-2 x+7=0$ $\begin{array}{ll} \Rightarrow \quad y =2 x-7 \\ \text { Slope } \quad(m) =2 \\ \therefore \text { Slope of } P Q =m_{P Q}=2 \\ \text { Equation of } P Q & \end{array}$ $(y-4)=2(x-3)$
On putting, the value of $y$ in Eq. (ii), we get $\begin{aligned} & & 3 x+2(2 x-2)+7 & =0 \\ & \Rightarrow & 7 x=-3 & \Rightarrow x=-\frac{3}{7} \end{aligned}$ Then, $y=2 \times\left(-\frac{3}{7}\right)-2=-\frac{20}{7}$ So, coordinates of $Q=\left(-\frac{3}{7}, \frac{-20}{7}\right)$ Thus, distance $P Q=\sqrt{\left(3-\left(\frac{-3}{7}\right)^2\right)+\left(4-\left(-\frac{20}{7}\right)\right)^2}=\frac{24 \sqrt{5}}{7}$

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