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# Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Q : 8         Find equation of the line perpendicular to the line $x-7y+5=0$ and having  $x$intercept $3$.

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It is given that line is  perpendicular to the line $x-7y+5=0$
we can rewrite it as
$y = \frac{x}{7}+\frac{5}{7}$
Slope of line $x-7y+5=0$   ( m' ) = $\frac{1}{7}$
Now,
The slope of the line is      $m = \frac{-1}{m'} = -7 \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
Now, the equation of the line with  $x$intercept $3$  i.e. (3, 0) and  with slope -7 is
$(y-0)=-7(x-3)$
$y = -7x+21$
$7x+y-21=0$

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