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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find equation of the line perpendicular to the line $x-7y+5=0$ and having $x$ intercept 3.

$\begin{array}{1 1}(A)\;7x-y=21 \\(B)\; 7x+y=-21 \\(C)\; y-7x+21=0 \\(D)\;7x+y=21 \end{array} $

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Toolbox:
  • (a) If two lines are perpendicular then the product of their slopes is -1. (i.e.,) $m_1m_2=-1$ where $m_1$ and $m_2$ are the slopes of the perpendicular lines.
  • (b) Equation of the line with slope $m$ and $x$ intercept $d$ is $ y=m(x-d)$
Given equation is $x-7y+5=0$
This can be written as
$ \qquad 7y=x+5$
$ \quad \Rightarrow y= \large\frac{1}{7}$$x+\large\frac{5}{7}$
This is of the form $y=mx+c$
Clearly slope of the line is $ \large\frac{1}{7}$
Hence the slope of the line which is perpendicular to this line is $ \qquad m = -\large\frac{1}{\Large\frac{1}{7}}$$=-7$
$\therefore$ The equation of the line with slope -7 and $x$ - intercept 3 is given by
$ \qquad y=m(x-d)$
$\qquad y=-7(x-3)$
$ \quad \Rightarrow y=-7x+21$
(i.e.,) $7x+y=21$ is the equation of the required line.
answered May 10, 2014 by thanvigandhi_1
 

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