Question

Which of the following pair of lines are parallel to each other?

• A. $x+7y=6$; $2x+14y=12$
• B. $5x+7y=12$; $2x+14y=12$
• C. $x+2y=7$; $3x+6y=28$
• D. $11x+2y=4$; $3x+6y=10$

Answer 1

The correct option is C $x+2y=7$; $3x+6y=28$

A pair of equations is said to form parallel lines if they have no solution.

Let two equations be $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$.

Then, the condition for a pair of linear equations to be parallel:
$\frac{a1}{a2}=\frac{b1}{b2}\ne \frac{c1}{c2}$

For equations:
$x+2y=7$; $3x+6y=21$
$a1=1;b1=2;c1=-7$
$a2=3;b2=6;c2=-28$

$\frac{1}{3}=\frac{2}{6}\ne \frac{-7}{-28}$
$\frac{1}{3}=\frac{1}{3}\ne \frac{1}{4}$

$\therefore$ $x+2y=7$; $3x+6y=28$ are parallel to each other.

For equations:
$x+7y=6$; $2x+14y=12$
$a1=1;b1=7;c1=-6$
$a2=2;b2=14;c2=-12$

$\frac{1}{2}=\frac{7}{14}\ne \frac{-6}{-12}$
$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$

$\therefore$ $x+2y=7$; $3x+6y=28$ are not parallel to each other.


For equations:
$5x+7y=12$; $2x+14y=12$
$a1=5;b1=7;c1=-12$
$a2=2;b2=14;c2=-12$

$\frac{5}{2}=\frac{7}{14}\ne \frac{-12}{-12}$
$\frac{5}{2}\ne \frac{1}{2}\ne \frac{1}{1}$

$\therefore$ $5x+7y=12$; $2x+14y=12$ are not parallel to each other.


For equations:
$11x+2y=4$; $3x+6y=10$
$a1=11;b1=2;c1=-4$
$a2=3;b2=6;c2=-10$

$\frac{11}{3}=\frac{2}{6}\ne \frac{-4}{-10}$
$\frac{11}{3}\ne \frac{1}{3}\ne \frac{2}{5}$

$\therefore$ $5x+7y=12$; $2x+14y=12$ are not parallel to each other.