Question

Find the value of 'k' for which the pair of equations 2x - ky + 3 = 0, 4x + 6y - 5 =0 represent parallel lines.

Answer 1

Dear Student,

Answer: k = -3

Solution:

if two lines are parallel that means they never meet at all,thus the condition of being no solution must apply.

Condition of No solution of pair of linear equation is given as

$\frac{a1}{a2} = \frac{b1}{b2} \: \: not \: = \: to \: \: \frac{c1}{c2}$
here all these are coefficient of x ,y and constant term respectively

2x - ky + 3 = 0,
here a1 = 2 , b1 = -k, c1 = 3

4x + 6y - 5 =0

a2 = 4 ,b2 = 6 ,c2 = -5

$\frac{2}{4} = \frac{ - k}{6} \: \: not \: equal \: to \: \frac{3}{ - 5} \\ \\ 2 \times 6 = - k \times 4 \\ \\ k = - 3$
or k ≠ 18/5

Hope it helps you.

Answer 2

Hi ,

Compare given equations

2x - ky + 3 = 0 ,

4x + 6y - 5 = 0 with

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

we get ,

a1 = 2 , b1 = -k , c1 = 3 ;

a2 = 4 , b2 = 6 , c2 = -5 ;

a1/a2 = b1/b2 ≠ c1/c2

[ Since , lines are parallel ]

a1/a2 = b1/b2

2/4 = ( -k )/6

( 2 × -6 )/4 = k

-3 = k

Therefore ,

k = -3

I hope this helps you.

: )