The distance of the point $(-2, 3)$ from the line $x - y - 5 = 0$ is

  • A
    $5\sqrt{2}$
  • B
    $2\sqrt{5}$
  • C
    $3\sqrt{5}$
  • D
    $5\sqrt{3}$

Explore More

Similar Questions

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

Find the distance of the point $(2, 3)$ from the line $2x - 3y + 9 = 0$ measured along the direction of the line $x - y + 1 = 0$.

Difficult
View Solution

Let $f(\theta)$ be the distance of the line $(\sqrt{\sin \theta})x + (\sqrt{\cos \theta})y + 1 = 0$ from the origin. Then the range of $f(\theta)$ is -

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side 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

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