Question

Find the value of 'K'for which pair of linear equations 3x-9y+7=0
and x+ky+5=0 represent parallel lines.​

Answer 1

$\huge{\mathcal{\blue{Answer}}}$

for parallel lines the condition is

a1/a2=b1/b2≠c1/c2

putting the values:

3/1=-9/k≠7/5

$\frac{3}{1} = \frac{ - 9}{k} \\ 3k = - 9 \\ k = \frac{ - 9}{3}$

$\huge{\mathcal{\blue{k=-3}}}$

Answer 2

Answer: -3

Step-by-step explanation:

Given lines : 3x-9y+7=0 and x+ky+5=0

Now

$a_1x+b_1y+c_1=0 \text { and } a_2x+b_2y+c_2=0 \text { then for parallel lines} \\\\\dfrac{a_1}{a_2}= \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

Therefore

$\dfrac{3}{1} =\dfrac{-9}{k} \ne \dfrac{7}{5} \\\\\therefore k= -3$

Hence, the value of k is -3