Question 1008437
Given a line with equation ____ ???.
The problem is broken, because we did not get the equation,
so I will fix it by making up an equation.
I will assume that your equation has an {{{x}}} AND a {{{y}}} .
If there is an {{{x}}} AND a {{{y}}} , your line has a slope, and so does the perpendicular line.
I will use equation {{{y=-2x+3}}} ,
which is given in slope-intercept form {{{y=mx+b}}}
The slope of that line is {{{m=-2}}} and the y-intercept is {{{b=3}}} .
The slope of a perpendicular line is {{{m[perp]=(-1)/m=(-1)/(-2)=1/2}}}
A line with slope {{{m}}} passing through point {{{P(x[P],y[P])}}}
can be written in point-slope form as
{{{y-y[P]=m(x-x[P])}}} .
For the perpendicular line in the answer to the problem as I fixed it,
the slope is {{{m=1/2}}} and the point is {{{P(2, -3)}}} with{{{system(x[P]=2,y[P]=-3)}}} , so the equation is
{{{y-(-3)=(1/2)(x-2)}}}<-->{{{highlight(y+3=(1/2)(x-2))}}} .