If $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$, then the lines $a_i x + b_i y + c_i = 0$ $(i = 1, 2, 3)$ represent:

  • A
    parallel lines if $\frac{a_i}{a_j} \neq \frac{b_i}{b_j} \neq \frac{c_i}{c_j}$ $(i \neq j)$
  • B
    coincident lines if $\frac{a_i}{a_j} = \frac{b_i}{b_j}$ $(i \neq j)$
  • C
    concurrent lines but not coincident if $\frac{a_i}{a_j} = \frac{b_i}{b_j} = \frac{c_i}{c_j}$ $(i \neq j)$
  • D
    concurrent lines if $\frac{a_i}{a_j} \neq \frac{b_i}{b_j} \neq \frac{c_i}{c_j}$ $(i \neq j)$

Explore More

Similar Questions

For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

The equation of the line parallel to the $y$-axis and passing through the point of intersection of the lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is:

If the lines $ax + y + 1 = 0, x + by + 1 = 0$ and $x + y + c = 0$ ($a, b, c$ being distinct and different from $1$) are concurrent,then $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

Difficult
View Solution

If $3a + 5b + 6c = 0$,then the family of lines $ax + by + c = 0$ passes through the fixed point:

If $x+2y-3=0$,$3x+4y-7=0$,$2x+3y-4=0$,and $4x+5y-6=0$ are the equations of four lines,then

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