Find the equation of the line perpendicular t | vedclass

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(A) The given equation of the line is $x-7y+5=0$.Rewriting in slope-intercept form $y=mx+c$,we get $7y=x+5$,or $y=\frac{1}{7}x+\frac{5}{7}$.Therefore,

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$7x+y=21$

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(A) The given equation of the line is $x-7y+5=0$.Rewriting in slope-intercept form $y=mx+c$,we get $7y=x+5$,or $y=\frac{1}{7}x+\frac{5}{7}$.Therefore,the slope of the given line is $m_1 = \frac{1}{7}$.The slope of the line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{1/7} = -7$.The equation of a line with slope $m$ and $x$-intercept $d$ is given by $y=m(x-d)$.Substituting $m=-7$ and $d=3$,we get $y=-7(x-3)$.$y = -7x + 21$.Rearranging the terms,we get $7x+y=21$.

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

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