Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are column c1 a. parallel to y axis is column c2 i. lambda =-3/4

The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are

Column C1 Column C2
a) parallel to y-axis is i) \lambda =-\frac{3}{4}
b) perpendicular to 7x+y-4=0 ii) \lambda =-\frac{1}{3}
c) Passes through (1,2) is iii) \lambda =-\frac{17}{41}
d) parallel to x-axis is iv) \lambda =3


 

Answers (1)

a ) Given equation is (2x+3y+4)+ λ(6x-y+12)=0  

  2x+3y+4+6λx-λy+12λ=0

   (2+6λ)x+(3-λ)y+4+12λ=0…. (i) 

 If equation i is parallel to y-axis, then 3-λ=0  

 λ=3   

 Hence, iv.    

b) Given equation is 2x+3y+4+λ6x-y+12=0   

 2x+3y+4+6λx-λy+12λ=0 

   (3-λ)y= -4-12λ-(2+6λ)x   

 y=-\left (\frac{2+6\lambda }{3-\lambda } \right )x+\left ( -1 \right )\left (\frac{4+12\lambda }{3-\lambda } \right )

 Since, the above equation is in y=mx+b  form  

 So the slope of equation (i)   is  m_{1}=-\left (\frac{2+6\lambda }{3-\lambda } \right )

 Now the second equation is 7x+y-4=0…..(ii)  

y=-7x+4

  So, the slope of equation (ii) is  m2=-7  

 Now equation i is perpendicular to equation (ii)  

 m1m2=-1 

-\left (\frac{2+6\lambda }{3-\lambda } \right )*\left ( -7 \right )=-1

   (2+6λ)*7=-(3-λ)   

 On solving we get λ=-17/41  

 Hence, (b)-(iii) 

  c) Given equation is (2x+3y+4)+λ(6x-y+12)=0  

 If the above equation passes through the point (1,2) then [2*1+3*2+4]+λ[6*1-2+12]=0    

  2+6+4+λ(6+10)=0   

  12+16λ=0  

  12=-16λ  

   λ=-12/16=-3/4  

 Hence,  (c)-(i)  

  d) Given equation is (2x+3y+4)+λ(6x-y+12)=0  

 2x+3y+4+6λx-λy+12λ=0  

  (2+6λ)x+(3-λ)y+4+12λ=0…..(i)  

  If equation (i) is parallel to x axis, then 2+6λ=0  6λ=-2 

  λ=-1/3   

 (d)-(ii) 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question