Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and Column C1 a) through the point (2,1) is Column C2 (i) 2x - y e=4

The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Column C1 Column C2
a) through the point (2,1) is i) 2x-y =4
b)perpendicular to the line x+2y+1=0 is ii)x+y-5=0                               
c)parallel to the line 3x+4y+5=0 is iii)x-y-1=0
d) equally inclined to the axes is iv) 3x-4y-1=0

 

Answers (1)

.a)  Given equations are 2x-3y=0….(i)  and 4x-5y=2….(ii )

 Equation of line passing through equations (i) and (ii), we get 2x-3y+λ4x-5y-2=0….(iii) 

   If the above equation passes through the point (2,1), we get (2*2-3*1)+λ(4*2-5*1-2)=0   

  (4-3)+λ(8-5-2)=0  

 1+λ=0  

 λ=-1  

  Putting the value of λ in equation iii, we get  (2x-3y)+(-1)(4x-5y-2)=0 

  2x-3y-4x+5y+2=0 

 -2x+2y+2=0  

 x-y-1=0 

  Hence, (a)-(iii)  

  b) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii)  

Equation of line passing through equation i and ii, we get 2x-3y+λ4x-5y-2=0…(iii) 

    2x-3y+4λx-5λy-2λ=0  

  x(2+4λ)-y(3+5λ)-2λ=0  

  -y(3+5λ)= -(2+4λ)+2λ  

   y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )+\left ( -\frac{2}{3+5\lambda } \right )

   Slope of equation iii is m1=\left ( \frac{2+4\lambda }{3+5\lambda } \right )

  Now, we find the slope of thegiven line x+2y+1=0…..(iv) 

  2y=-x-1   y=-1/2x+(-1/2)

 So slope of equation (iv) is m2=(-1/2)

  We know that, if two lines are perpendicular to each other then the product of their slopes is  equal to-1 

  So,   m1*m2=-1   

\left ( \frac{2+4\lambda }{3+5\lambda } \right )*\left (-\frac{1}{2} \right )=-1

   2+4λ=2*(3+5λ) 

 2+4λ=6+10λ  

-6λ=4 

 λ=-4/6= -2/3  

  Putting the value of λ in equation (iii) we get   (2x-3y)+(-2/3)(4x-5y-2)=0  

  6x-9y-8x+10y+4=0  

 -2x+y+4=0  

 2x-y=4  

b-i

   c)  Given equations are 2x-3y=0….(i)  and 4x-5y=2…..(ii)

   Equation of line passing through equation iand (ii), we get  (2x-3y)+ λ(4x-5y-2)=0….iii 

2x-3y+4λx-5λy-2λ=0  

  x(2+4λ)-y(3+5λ)-2λ=0   

  y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )x+\left ( -\frac{2}{3+5\lambda } \right )

 So slope of equation iii is m1= \left ( \frac{2+4\lambda }{3+5\lambda } \right )

  Now, we find the slope of the given line 3x-4y+5=0  

 3x+5=4y    y=3/4x+5/4  

 So slope of equation is 3/4

  If 2 lines are parallel then their slopes are also equal  So \left ( \frac{2+4\lambda }{3+5\lambda } \right )=\frac{3}{4}

  4(2+4λ)=3(3+5λ) 

  8+16λ=9+15λ   

  λ=1  

 Putting the values of λ in equation iii, we get (2x-3y)+(4x-5y-2)=0  

   2x-3y+4x-5y-2=0   

 6x-8y-2=0 

  3x-4y-1=0  

 Hence, c-iv  

 d) Given equations are 2x-3y=0…(i)   and 4x-5y=2….(ii) 

Equation of line passing through equation (i) and (ii), we get  (2x-3y)+λ(4x-5y-2)=0  

  x(2+4λ)-y(3+5λ)-2λ=0   

-y(3+5λ)= -(2+4λ)+2λ 

y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )x+\left (\frac{-2 }{3+5\lambda } \right )

  Slope of equation iii is \left ( \frac{2+4\lambda }{3+5\lambda } \right )

Since the equation is equally inclined with axes, it means that the line makes equal angles with both the

 coordinate axes.

 It will have an angle of 450 or 1350   

 m2=tan450   and tan1350    

= 1 and-1 

  y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )=-1 \, \, or\, \, \left (\frac{2+4\lambda }{3+5\lambda } \right )=1

  2+4λ=-3+5λ or  2+4λ=3+5λ  

  So λ=-5/9  or-1

 Putting the value in equation iii we get (2x-3y)+(-5/9)(4x-5y-)2=0     

 18x-27y-20x+25y+10=0 

 -2x-2y+10=0  

x+y-5=0 

  If λ=-1, then the required equation is  (2x-3y)+(-1)( 4x-5y-2)=0  

 2x-3y-4x+5y+2=0  

-2x+2y+2=0

  x-y-1=0   

d-ii, iii

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question