Question

If the lines given by $3x+2ky=2\mathrm{and}2\mathrm{x}+5\mathrm{y}+1=0$ are parallel, then the value of k is

(a) $\frac{-5}{4}$
(b) $\frac{2}{5}$
(c) $\frac{3}{2}$
(d) $\frac{15}{4}$

Answer 1

The correct option is (d).

The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁ = 2k, c₁ = −2 and a₂ = 2, b₂ = 5 and c₂ = 1
∴ $\frac{a_1}{a_2}=\frac{3}{2},\frac{b_1}{b_2}=\frac{2k}{5}\mathrm{and}\frac{c_1}{c_2}=\frac{-2}{1}$
For parallel lines, we have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
∴ $\frac{3}{2}=\frac{2k}{5}\ne \frac{-2}{1}$
⇒ $k=\frac{15}{4}$