Question

Pair of lines represented by equation 2x+y+3=0 and 4x+ky+6=0 will be parallel if value of k is

Answer 1

Question:

Pair of lines represented by the equations 2x + y + 3 = 0 and

4x + ky + 6 = 0 will be parallel

if k is .

Answer:

No real value.

Note:

If we consider a pair of linear equations in two variables say;

a1x + b1y + c1 = 0 and

a2x + b2y + c2 = 0.

Then,

The condition for which the two lines are parallel is given as;

a1/a2 = b1/b2 ≠ c1/c2

Solution:

Here,

The given pair of linear equations is;

2x + y + 3 = 0 and

4x + ky + 6 = 0.

Clearly,

Here , we have;

a1 = 2

a2 = 4

b1 = 1

b2 = k

c1 = 3

c2 = 6

Thus,

The condition for which the given lines

will be parallel, is given as;

=> a1/a2 = b1/b2 ≠ c1/c2

=> 2/4 = 1/k ≠ 3/6

=> 1/2 = 1/k ≠ 1/2

=> 2 = k ≠ 2

{ From above equation, we are getting that , k=2 and k≠2 simultaneously, which is not possible. }

Thus,

There exist no real value of k for which the given pair of lines are parallel to each other.

Conclusion;

The given pair of lines will be either coincidence or intersecting depending on the value of k.

Moreover,

If k = 2 ,then the given pair of lines will be coincidence otherwise for other real values of k ( ie, if k ≠ 2) they will be interesting.