If the lines given by 3x + 2ky = 2 and 3x + 5y = 1 are parallel, then find the value of 'k'.

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Solution

Condition for parallel lines is
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Given 3x + 2ky – 2 = 0
And 2x + 5y – 1 = 0
Here, $a 1 = 3, b 1 = 2 k, c 1 = - 2$
and $a_{2} = 2, b_{2} = 5, c_{2} = - 1$

From Eq (i), $\frac{3}{2} = \frac{2 k}{5}$
$∴ k = \frac{15}{4}$

Also, $\frac{c_{1}}{c_{2}} = \frac{- 2}{- 1} = 2$

Thus, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

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Question 7
If the lines given by 3x + 2ky = 2 and 3x + 5y = 1 are parallel, then the value of k is
(a) $- \frac{5}{4}$
(b) $\frac{1}{2}$
(c) $\frac{15}{4}$
(d) $\frac{3}{2}$

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If the lines given by 3x + 2ky = 2 and 3x + 5y = 1 are parallel, then find the value of 'k'.

Condition for parallel lines is a1a2=b1b2≠c1c2 Given 3x + 2ky – 2 = 0 And 2x + 5y – 1 = 0 Here, a1=3, b1=2k, c1=−2 and a2=2, b2=5, c2=−1 From Eq (i), 32=2k5 ∴k=154 Also, c1c2=−2−1=2 Thus, a1a2=b1b2≠c1c2