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For example, if I need to find the equation of the line parallel to $$2x-3y=4$$ which passes through the point $(1,-5)$ I know how to do this by putting it into slope-intercept form first to find the slope and then plugging in the point to find the y-intercept.

Same thing for finding a line perpendicular to that point. I just wanted to know if I could do this without changing it into slope intercept form first

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The slope of this line is ${2\over 3}$ so the slope of perpendicular is $-{3\over 2}$ so the equation of perpenicular is $$y-(-5)= -{3\over 2}(x-1)$$

and the line parallel has the same slope, so ${2\over 3}$ an thus it equation is $$y-(-5)= {2\over 3}(x-1)$$

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  • $\begingroup$ thank you! so then I would have $y+5 = 2/3(x-1)$ but what about if i wanted it in the same for that it was given on top. so i want the answer to be in the form $2x-3y=4$. I know I can manipulate what I have to get it there but is there a way to get there directly? $\endgroup$
    – user130306
    May 18 '19 at 17:31
  • $\begingroup$ just get rid of denumerators and move variables to the left $\endgroup$
    – Aqua
    May 18 '19 at 17:32
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Simply substitute $x=1$ and $y=-5$ in $2x-5y$ to get $17$. Hence the equation of the line parallel to $2x-3y=4$ passing through $(1,-5)$ is $2x-5y=17$.

For the perpendicular line plug in the coordinates $(1,-5)$ in $3x+2y$ to get $3x+2y=-7$.

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