The distance between the lines $3x - 2y = 1$ and $6x - 4y + 9 = 0$ is

  • A
    $\frac{1}{\sqrt{52}}$
  • B
    $\frac{11}{\sqrt{52}}$
  • C
    $\frac{4}{\sqrt{13}}$
  • D
    $\frac{6}{\sqrt{13}}$

Explore More

Similar Questions

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

Find the distance between the parallel lines $l(x + y) + p = 0$ and $l(x + y) - r = 0$.

The number of lines which pass through the point $(2, -3)$ and are at a distance of $8$ from the point $(-1, 2)$ is:

If the points $(1, 2)$ and $(3, 4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x \cos \theta - y \sin \theta = k \cos 2 \theta$ and $x \sec \theta + y \csc \theta = k$ respectively,prove that $p^{2} + 4q^{2} = k^{2}$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo