Question

If the lines 3x+2ky – 2 = 0 and 2x+5y+1 = 0 are parallel, then what is the value of k?​

Answer 1

• The value of k = 15/4

Step-by-step explanation:

Given:

The lines 3x + 2ky - 2 = 0 and 2x + 5y + 1 = 0 are parallel.

To find:

The value of k for which the given equations satisfy they have parallel lines.

Now,

3x + 2ky - 2 = 0 --------(i)

2x + 5y + 1 = 0  --------(ii)

Here,

a₁ = 3 , b₁ = 2k , and c₁ = (-2)

a₂ = 2 , b₂ = 5 , and c₂ = 1

• The condition for parallel line

$\boxed{\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \ne \frac{c_{1}}{c_{2}}}$

So,

$\implies \frac{3}{2}=\frac{2k}{5} \ne \frac{-2}{1}$

⇒ 3/2 = (2k)/5

⇒ 5 * (3) = 2k * (2)

⇒ 15 = 4k

⇒ 25/4 = k

Thus,

$\boxed{k = \frac{15}{4}}$

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Verification:

Putting the value of k

⇒ b₁/b₂= 2k/5

= [ 2 *(15/4) ]/5

= (15/2)/5

= (15/2) * 1/5

= 3/2 = a₁/a₂

Hence,

• We got a₁/a₂ = b₁/b₂ verified.

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More Information,

• The parallel line has no solution as they will not intersect each other at any point. Their system if consistent.

Answer 2

Given :-

$\longrightarrow \: \sf3x + 2ky \times 2 = 0 \\ \\ \longrightarrow \: \sf2x + 5y + 1 = 0$

To Find :-

$\longrightarrow \: \sf Value \: of \: k$

Solution :-

Given line are parallel and condition for parallel line is

$\longrightarrow \: \sf \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \\ \\ \longrightarrow \: \sf \frac{a_1}{a_2} = \frac{b_1}{b_2}$

Here

$\longrightarrow \: \sf a_1 = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a_2 =2k \\ \\\longrightarrow \: \sf b_1 = 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b_2 =5$

Substitute values in formula

$\longrightarrow \: \sf \frac{3}{2k} = \frac{2}{5} \\ \\\longrightarrow \: \sf 2 \times 2k = 3 \times 5 \\ \\\longrightarrow \: \sf 4k = 15 \\ \\\longrightarrow \: \underline{ \boxed{ \sf k = \frac{15}{4} }}$