Question

The value of $k$, for which the system of equations $kx-3y+6=0$, $4x-6y+150$ represent parallel lines is _________.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

For a system of equations representing parallel lines means they do not intersect, hence, have no solutions.

For a system of equations $a_{1}x+b_{1}y+c_1=0;a_{2}x+b_{2}y+c_2=0$ to have no solution, the condition to be satisfied is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\Rightarrow \frac{k}{4}=\frac{-3}{-6}\ne \frac{6}{150}$

$\therefore$ For $k=2$, the given system of equation has no solution, hence represent parallel lines.