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Straight Lines — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSStraight Lines5 Marks
Question
Match the questions given under Column C1 with their appropriate answers given under the Column C2: The value of the λ, if the lines (2x + 3y + 4) + λ (6x - y + 12) = 0 are:
Column C1 Column C2
(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

Column C1 Column C2
(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) $\lambda=3$
Solution:
  1. 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)
  1. 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)
  1. 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)
  1. 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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