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
$\endgroup$2 Answers
$\begingroup$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 $$y-(-5)= {2\over 3}(x-1)$$
$\endgroup$ 2 $\begingroup$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 $3x+2y=-7$.
$\endgroup$
