Question 802591
The altitude through point R is the line that passes through R and is perpendicular to the line QS.
 
The slope of the line QS is
{{{(y[S]-y[Q])/(x[S]-x[Q])=(-3-(-1))/(5-(-4))=(-3+1)/(5+4)=-2/9}}}.
 
The product of the slopes of two perpendicular lines is {{{-1}}}.
So the slope of a line perpendicular to QS is
{{{m=(-1)/((-2/9))=(-1)*(-9/2)=9/2}}}
 
A point, {{{ "( x , y )" }}}, on a line with slope {{{m}}} that passes through {{{ P ( x[P] , y[P] ) }}} satisfies
{{{m=(y-y[P])/(x-x[P])}}} <--> {{{y-y[P]=m(x-x[P])}}}
(The equation {{{y-y[P]=m(x-x[P])}}} is called the point-slope form of the equation of the line).
Applying that to point {{{ R ( -1 , 3 ) }}} and slope {{{m=9/2}}} we get
{{{y-3=(9/2)(x-(-1))}}} --> {{{y-3=(9/2)(x+1)}}} --> {{{y-3=(9/2)x+9/2}}} --> {{{y=(9/2)x+9/2+3}}} --> {{{highlight(y=(9/2)x+15/2)}}}
{{{drawing(360,300,-5,7,-5,5,
grid(1),
blue(line(-4,-1,-1,3)),
blue(line(-1,3,5,-3)),
blue(line(-4,-1,5,-3)),
green(line(-5,-15,1,12)),
green(line(-1.9,-1.05,-1.45,-1.15)),
green(line(-1.45,-1.15,-1.55,-1.6)),
locate(-4.3,-1,Q),locate(-1.3,3.5,R),locate(5.1,-3,S)
)}}}