Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is (A)-5by4 (B)2by5 (C)15by4 (D)3by2

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A)\frac{-5}{4}
(B)\frac{2}{5}
(C)\frac{15}{4}|
(D)\frac{3}{2}
 

Answers (1)

Answer: C
Solution:
Equations are
3x + 2ky = 2
2x + 5y + 1 = 0
In equation
3x + 2ky – 2 = 0
a1 = 3, b1 = 2k, c1 = –2
In equation
2x + 5y + 1 = 0
a2 = 2, b2 = 5, c2 = 1
\frac{a_{1}}{a_{2}}= \frac{3}{2}
\frac{b_{1}}{b_{2}}= \frac{2k}{5}
\frac{c_{1}}{c_{2}}= \frac{-2}{1}=-2
Since lines are parallel hence
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}

\frac{3}{2}=\frac{2k}{5}\neq-2           ……(1)

\frac{3}{2}=\frac{2k}{5}         ( from equation (1))

2k=\frac{15}{2}\Rightarrow k=\frac{15}{4}

Hence the value of k is \frac{15}{4} .

Posted by

infoexpert27

View full answer

Similar Questions

Latest Question