Question

for which value of k will the line represented buy the following pair of linear equation be parallel
x-2y+3=0
3×-6y+k=0​

Answer 1

Solution -

Let the two lines be -

x-2y+3=0

And,

3x-6y+k=0

Now,

=>x-2y+3=0

=> x+(-2)y+3=0

Here,

$a_1 = 1$

$b_1 = - 2$

$c_1 = 3$

And,

=>3x-6y+k=0

=> 3x+(-6)y+k=0

Here,

$a_2 = 3$

$b_2 = - 6$

$c_2 = k$

Since, two lines are parallel,

Therefore,

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

Putting the values we will get -

$\implies \frac{1}{3} = \frac{ - 2}{ - 6} \neq \frac{3}{k}$

$\implies\frac{1}{3} = \frac{1}{3} \neq \frac{3}{k}$

$\implies \: \frac{1}{3} \neq \frac{3}{k}$

By cross multiplying -

$\implies \: k\neq 3 \times 3$

$\implies \: k\neq9$

Therefore the value of k will be any real values except 9