Q74 / 200 · Manipal (MET) 2014 · Mathematics

The distance of the point $(3,5)$ from $2 x+3 y-14=0$ measured parallel to $x-2 y=1$ is

  • A
    $\frac{7}{\sqrt{5}}$
  • B
    $\frac{7}{\sqrt{13}}$
  • Correct answer
    C
    $\sqrt{5}$
  • D
    $\sqrt{13}$

Solution

Let the equation of the line parallel to $x-2 y=1$ is $x-2 y+\lambda=0$ Since, it passes through $(3,5)$ $\therefore \quad 3-10+\lambda=0 \Rightarrow \lambda=7$ $\therefore$ The line is $x-2 y+7=0$. The point of intersection of $x-2 y+7=0$ and $2 x+3 y-14=0$ is $(1,4)$. $\therefore$ The distance between $(3,5)$ and $(1,4)$ $=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{4+1}=\sqrt{5}$

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

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