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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  xintercept 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  xintercept 3  i.e. (3, 0) and  with slope -7 is 
(y-0)=-7(x-3)
y = -7x+21
7x+y-21=0

Posted by

Gautam harsolia

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