On which side of the line $3x - 4y + 5 = 0$ does the point $(3, -4)$ lie?

  • A
    Perpendicular distance
  • B
    Origin side
  • C
    Opposite side
  • D
    Adjacent side

Explore More

Similar Questions

The distance of the point of intersection of the lines $2x - 3y + 5 = 0$ and $3x + 4y = 0$ from the line $5x - 2y = 0$ is

The perpendicular distance of the straight line $12x + 5y = 7$ from the origin is equal to

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then $\frac{1}{a^2}+\frac{1}{b^2}=$

What is the position of the point $(8, -9)$ with respect to the lines $2x + 3y - 4 = 0$ and $6x + 9y + 8 = 0$?

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