Question 1008239
The slope of {{{AC}}} is
{{{(y[C]-y[A])/(x[C]-x[A])=(6-2)/(3-(-1))=4/4=red(1)}}} ,
and the equation of  line {{{AC}}} in in point-slope form is
{{{y-y[A]=red(1)(x-x[A])}}}-->{{{y-(-1)=red(1)(x-2)}}}
We can convert to the slope-intercept form by solving for {{{y}}} ;
{{{y-(-1)=red(1)(x-2)}}}-->{{{y+1=x-2}}}-->{{{y=x-2-1}}}-->{{{y=x-3}}} .
The altitude to AC is the perpendicular segment from {{{B}}} to line {{{AC}}} ,
so its length is the distance from point {{{B}}} to line {{{AC}}}.
There is somewhere an ugly formula to calculate distance from a point to a line,
so we could plug numbers into that formula and be done.
 
Otherwise, since the altitude to {{{AC}}} is perpendicular to {{{AC}}} ,
the slope of that altitude is {{{-1/red(1)=-1}}} .
The altitude to {{{AC}}} is part of a line
with slope {{{-1}}} that contains point {{{B(-1,1)}}} .
The equation of that line in point-slope form is
{{{y-y[B]=-1(x-x[B])}}}-->{{{y-1=-1(x-(-1))}}} .
We can convert to the slope-intercept form by solving for {{{y}}} ;
{{{y-1=-1(x-(-1))}}}-->{{{y-1=-1(x+1)}}}-->{{{y-1=-x-1)}}}-->{{{y=-x}}} .
We can find point {{{D}}} at the intersection of the two perpendicular lines
{{{system(y=x-3,y=-x)}}}-->{{{system(-x=x-3,y=-x)}}}-->{{{system(-2x=-3,y=-x)}}}-->{{{system(x=(-3)/(-2),y=-x)}}}-->{{{system(x=1.5,y=-x)}}}-->{{{system(x=1.5,y=-1.5)}}} .
So the altitude is {{{BD}}} with {{{D(1.5,-1.5)}}} ,
and the length of altitude {{{BD}}} is
{{{sqrt((x[D]-x[B])^2+(y[D]-y[B])^2)=sqrt((1.5-(-1))^2+(-1.5-1)^2)=sqrt((1.5+1)^2+(-2.5)^2)=sqrt(2.5^2+2.5^2)=sqrt(2.5^2*2)=highlight(2.5sqrt(2))}}}