Match the questions given under Column C_1 with their appropriate answers given under the Column C_2: The value of the , if the lines (2x + 3y + 4) + (6x - y + 12) = 0 are: Column C_1 Column C_2 (a) Parallel to y-axis is (i) =-3 4 (b) Perpendicular to 7x + y - 4 = 0 is (ii) =-1 3 (c) Passes through (1, 2) is (iii) =-17 41 (d) Parallel to x axis is (iv) =3

Column C_1 Column C_2 (a) Parallel to y-axis is (i) =-3 4 (b) Perpendicular to 7x + y - 4 = 0 is (ii) =-1 3 (c) Passes through (1, 2) is (iii) =-17 41 (d) Parallel to x axis is (iv) =3 Given equcation is (2x+3y+4)+ (6x- y+12)=0 (2+6 )x+(3- )y+4+12 =0 ...(i) If eq.(i) is parallel to y-axis, then 3- =0 =3 Hence, (a) ⇔ (iv) Given lines are (2x+3y+4)+ (6x-y+12)=0 ..(i) (2+6 )x+3(3- )y+12 =0 Slope =- (2+6 3- ) Second equation is 7x+y-4=0 ...(ii) Slope = -7 If eq. (i) eq. (ii) are perpendilcular to each other (-7) [- (2+6 3- ) ]=1 14+42 3- = -1 41 =-17 =-17 41 Hence, (b) ⇔ (iii) Given equation is (2x+3y+4)+ (6x- y+12)=0 ...(i) If eq.(i) passes throgh the given ponit (1, 2) then (2 1+3 2+4) + (6 1-0+12)=0 (2+6+4) + (6-2+12)=0 12+16 =0 =-12 16 =-3 4 Hence, (c) ⇔ (i) The equation is (2x+3y+4)+ (6x- y+12)=0 (2x+3y+4)+ (6x- y+12)=0 If eq. (i) is parallel to x-axis, then 2+6 =-1 3 Hence, (d) ⇔ (ii)

Match the questions given under Column C_1 with their appropriate…

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Question

Match the questions given under $\text{Column}\ C_1$ with their appropriate answers given under the $\text{Column}\ C_2$: The value of the $\lambda ,$ if the lines $(2x + 3y + 4) + \lambda (6x - y + 12) = 0$ are:

	$\text{Column}\ C_1$		$\text{Column}\ C_2$
$(a)$	Parallel to $y-$axis is	$(i)$	$\lambda=-\frac{3}{4}$
$(b)$	Perpendicular to $7x + y - 4 = 0$ is	$(ii)$	$\lambda=-\frac{1}{3}$
$(c)$	Passes through $(1, 2)$ is	$(iii)$	$\lambda=-\frac{17}{41}$
$(d)$	Parallel to $x$ axis is	$(iv)$	$\lambda=3$

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Answer

	$\text{Column}\ C_1$		$\text{Column}\ C_2$
$(a)$	Parallel to $y-$axis is	$(i)$	$\lambda=-\frac{3}{4}$
$(b)$	Perpendicular to $7x + y - 4 = 0$ is	$(ii)$	$\lambda=-\frac{1}{3}$
$(c)$	Passes through $(1, 2)$ is	$(iii)$	$\lambda=-\frac{17}{41}$
$(d)$	Parallel to $x$ axis is	$(iv)$	$\lambda=3$

Given equcation is
$(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$
$\Rightarrow(2+6\lambda)\text{x}+(3-\lambda)\text{y}+4+12\lambda=0\ \ ...\text{(i)}$ If $eq.(i)$ is parallel to $y-$axis,
then $3- \lambda=0$
$\Rightarrow\lambda=3$
Hence$, (a)$ ⇔ $(iv)$
Given lines are
$(2\text{x}+3\text{y}+4)+\lambda(6\text{x}-\text{y}+12)=0 ..(i)$
$\Rightarrow(2+6\lambda)\text{x}+3(3-\lambda)\text{y}+12\lambda=0$ Slope $=-\bigg(\frac{2+6\lambda}{3-\lambda}\bigg)$ Second equation is $7\text{x}+\text{y}-4=0\ \ ...\text{(ii)}$ Slope $= -7$ If $eq. (i)\ eq. (ii)$ are perpendilcular to each other
$\therefore(-7)\bigg[-\bigg(\frac{2+6\lambda}{3-\lambda}\bigg)\bigg]=1$
$\Rightarrow\frac{14+42\lambda}{3-\lambda} = -1$
$41\lambda=-17$
$\lambda=-\frac{17}{41}$
Hence$, (b)$ ⇔ $(iii)$
Given equation is $(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0\ \ ...\text{(i)}$
If $eq.(i)$ passes throgh the given ponit $(1, 2)$ then $(2\times1+3\times2+4) +\lambda(6\times1-0+12)=0$
$\Rightarrow(2+6+4) +\lambda(6-2+12)=0$
$\Rightarrow 12+16\lambda=0$
$\Rightarrow=\frac{-12}{16}=\frac{-3}{4}$
Hence$, (c)$ ⇔ $(i)$
The equation is $(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$
$\Rightarrow(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$ If $eq. (i)$ is parallel to $x-$axis$,$ then $2+6\lambda$
$\Rightarrow\lambda=\frac{-1}{3}$
Hence$, (d)$ ⇔ $(ii)$