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# Match the question given under Column $C_1$ with their appropriate answer given under the Column $C_2$ for the following : The value of the $\lambda$, if the lines $(2x + 3y + 4) +\lambda (6x – y + 12) = 0$ are

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

Toolbox:
• Equation of a line parallel to y - axis is x = a constant
• Slope of a line is $-\large\frac{coefficient \: of \: x }{coefficient \: of \: y}$
Step 1 :
Equation of line can be written as $x(2+ 6\lambda )+y(3-\lambda )+4+12\lambda = 0$-----------(1)
If this line is parallel to y - axis then equation of the line is x = a constant.
$\Rightarrow (3- \lambda )=0$
$\therefore \lambda = 3$
(a) Parallel to y - axis is $\qquad$ $\lambda = 3$
(i.e) Column $C_1$ - a matches Column $C_2$ - (iv)
Slope of the line (1) is $-\large\frac{(2+6\lambda)}{(3- \lambda )}$
If this is perpendicular to the line $7x+y-4=0$ whose slope is $- \large\frac{7}{1}$$=-7 then -\large\frac{(2+6\lambda)}{3-\lambda}$$ \times -7=-1$
$\Rightarrow 7(2+6\lambda )= - (3- \lambda)$
$\Rightarrow 14+42\lambda = -3+ \lambda$
$\Rightarrow 41 \lambda = -17$
$\therefore \lambda= -\large\frac{17}{41}$
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(b) Perpendicular to 7x+y-4=0 $\qquad$ $\lambda = -\large\frac{17}{41}$
(i.e) Coulmn $C_1$ - b matches Coulmn $C_2$ - (iii)
If the line passes through (1,2) then
$[ 2(1) +3(2)+4] + \lambda [ 6(1)-2+12]=0$
$\Rightarrow 12+16 \lambda = 0$
$\therefore \lambda = -\large\frac{12}{16}$$=-\large\frac{3}{4}$
(c) passes through (1,2) is $\qquad \lambda = - \large\frac{3}{4}$
(i.e) Coulmn $C_1$ - c matches Column $C_2$ - (i)
Parallel to x - axis , then
$2+6\lambda = 0$
$\Rightarrow 6\lambda = -2$
$\lambda = -\large\frac{1}{3}$
(d) parallel to x axis $\qquad \lambda = - \large\frac{1}{3}$
(i.e) Coulmn $C_1$ - d matches Column $C_2$ - (ii)
edited Jul 9, 2014