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Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsStraight Lines5 Marks
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 $(iv)$ $\lambda=3$
✓

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$, (d)$ ⇔ $(ii)$

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