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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the line parallel to $y$ - axis and drawn through the point of intersection of the lines $x – 7y + 5 = 0$ and $3x + y = \theta.$

$\begin {array} {1 1} (A)\;x = \large\frac{5}{22} & \quad (B)\;x=-\large\frac{5}{22} \\ (C)\;y = \large\frac{5}{22} & \quad (D)\;y =- \large\frac{5}{22} \end {array}$

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  • Equation of a line passing parallel to $y$ - axis is $x=a$, where $a$ is a constant.
Given equation of the two lines are
$ \qquad x-7y+5=0$
$ \qquad 3x+y=0$
On solving the two equations for $x$ and $y$ we get,
$ \qquad x-7(-3x)+5=0$
$ \qquad 22x+5=0 \Rightarrow x = - \large\frac{5}{22}$
and $y = \large\frac{15}{22}$
The point of intersection of the above two lines is $ \bigg( -\large\frac{5}{22}$$, \large\frac{15}{22} \bigg)$
Since the line $x=a$ passes through the point $ \bigg( -\large\frac{5}{22}$$, \large\frac{15}{22} \bigg)$
$ a = -\large\frac{5}{22}$
Hence the required equation of the line is
$ x = -\large\frac{5}{22}$
answered May 20, 2014 by thanvigandhi_1

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