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NCERT Exemplar · Q58

Q.The value of λ\lambda, if the lines (2x+3y+4)+λ(6x−y+12)=0(2x+3y+4)+\lambda(6x-y+12)=0 are — match Column C1C_1 with Column C2C_2. Column C1C_1:

(a) parallel to yy-axis is;
(b) perpendicular to 7x+y−4=07x+y-4=0 is;
(c) passes through (1,2)(1,2) is;
(d) parallel to xx axis is. Column C2C_2:
(i) λ=−34\lambda=-\dfrac{3}{4};
(ii) λ=−13\lambda=-\dfrac{1}{3};
(iii) λ=−1741\lambda=-\dfrac{17}{41};
(iv) λ=3\lambda=3.
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For the family (2+6λ)x+(3−λ)y+(4+12λ)=0(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0: (a)→\to(iv) λ=3\lambda=3; (b)→\to(iii) λ=−1741\lambda=-\tfrac{17}{41}; (c)→\to(i) λ=−34\lambda=-\tfrac34; (d)→\to(ii) λ=−13\lambda=-\tfrac13.

Solution

Rewrite the family of lines as

(2+6λ)x+(3−λ)y+(4+12λ)=0,slope m=−2+6λ3−λ.(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0,\qquad \text{slope }m=-\frac{2+6\lambda}{3-\lambda}.

(a) Parallel to the yy-axis (vertical line, so no yy-term): 3−λ=0⇒λ=3.3-\lambda=0\Rightarrow\lambda=3. →\to (iv)

(b) Perpendicular to 7x+y−4=07x+y-4=0 (its slope is −7-7, so the required slope is 17\tfrac17):

−2+6λ3−λ=17⇒−7(2+6λ)=3−λ⇒−14−42λ=3−λ⇒λ=−1741.-\frac{2+6\lambda}{3-\lambda}=\frac17\Rightarrow -7(2+6\lambda)=3-\lambda\Rightarrow -14-42\lambda=3-\lambda\Rightarrow \lambda=-\frac{17}{41}. →\to (iii)

(c) Passes through (1,2)(1,2): …

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