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

Column C1                                                                  Column C2

i.       Parallel to y-axis is                                                 (i) λ = - 3 4  

ii.    Perpendicular to   7 x + y – 4 = 0 is                           (ii) λ = - 1 3  

iii.   passes through (1, 2) is                                            (iii) λ = - 17 41  

iv.  parallel to x-axis is                                                  (iv) λ = 3  

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This is a True or False Type Questions as classified in NCERT Exemplar

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(a)Given??  equation  is             (2x+3y+4)+λ(6x−y+12)=0⇒       (2+6λ)x+(3−λ)y+4+12λ=0                          …(i)       If  eqn.(i)  is  parallel  to  yaxis,  then             3−λ=0     ⇒λ=3       Hence,  (a)↔(iv)(b)Given??  lines  are             (2x+3y+4)+λ(6x−y+12)=0                           …(i)⇒       (2+6λ)x+(3−λ)y+4+12λ=0        Slope=−(2+6λ3−λ)          equation  is  7x+y−4=0                               …(ii)        Slope=−7       If  eqn.(i)  and  eqn.(ii)  are  perpendicular  to  each  other∴          (−7)[−(2+6λ3−λ)]=−1⇒                                14+42λ3−λ=−1⇒                                 14+42λ=−3+λ⇒                                   42λ−λ=−3−14⇒                                            41λ=−17⇒                                                 λ=−1741       Hence,  (b)↔(iii)(c)Given??  equation  is  (2x+3y+4)+λ(6x−y+12)=0                               …(i)       If  eqn.(i)    through  the  given    (1,2)  then       (2*1+3*2+4)+λ(6*1−2+12)=0⇒                     (2+6+4)+λ(6−2+12)=0⇒                                                        12+16λ=0⇒                                                                       λ=−1216=−34       Hence,  (c)↔(i)(d)The  given??  equation  is  (2x+3y+4)+λ(6x−y+12)=0⇒       (2+6λ)x+(3−λ)y+4+12λ=0                          …(i)       If  eqn.(i)  is  parallel  to  xaxis,  then             2+6λ=0     ⇒λ=−13       Hence,  (d)↔(ii)

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Maths NCERT Exemplar Solutions Class 11th Chapter Ten 2025

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