Question 1181232
Given triangle {{{ABC}}} with vertices 

A({{{6}}}, {{{3}}}), 
B({{{-5}}}, {{{-8}}}), and 
C({{{-3}}}, {{{6}}}), 


let BP and CQ be the perpendiculars dropped from B and C to their opposite sides of triangle ABC

a line through {{{BP }}} is perpendicular to a line through {{{AC}}}

slope of a line {{{y=mx+b}}} through {{{AC}}} is

{{{m=(6-3)/(-3-6)=-3/9=-1/3}}}

{{{y=-(1/3)x+b}}}.......use one point to find {{{b}}}

{{{3=-(1/3)6+b}}}

{{{3=-2+b}}}

{{{b=5}}}

a line through {{{AC}}} is:{{{y=-(1/3)x+5}}}


then a perpendicular line {{{BP }}} will have a slope {{{-1/(-1/3)=3}}}

{{{y=3x+b}}}........use point B

{{{-8=3(-5)+b}}}
{{{-8=-15+b}}}
{{{15-8=b}}}
{{{b=7}}}

and a perpendicular line {{{BP }}} is {{{y=3x+7}}}


now we need a line that passes through {{{CQ}}} which is perpendicular to BA

line BA have a slope {{{m=(-8-3)/(-5-6)=-11/-11=1}}}

{{{y=1x+b}}}......use one point to find {{{b}}}

{{{3=1*6+b}}}
{{{3-6=b}}}
{{{b=-3}}}

a line that passes through {{{BA}}} {{{y=x-3}}}  

a line that passes through {{{CQ}}} will have a slope {{{-1/1=-1}}}

{{{y=-x+b}}}....use coordinates of C
{{{6=-(-3)+b}}}
{{{6=3+b}}}
{{{b=3}}}
a line that passes through {{{CQ}}} is {{{y=-x+3}}}



the orthocenter {{{H}}} is intersection point of a lines that passes through  {{{BP }}} and {{{CQ}}}

 {{{y=3x+7}}}
{{{y=-x+3}}}
--------------------
{{{3x+7=-x+3}}}
{{{3x+x=3-7}}}
{{{4x=-4}}}
{{{x=-1}}}

 {{{y=3(-1)+7}}}
 {{{y=-3+7}}}
 {{{y=4}}}

  the coordinates of orthocenter H of triangle ABC are: ({{{-1}}},{{{4}}})


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(6,3,.12), locate(6,3,A),
circle(-5,-8,.12), locate(-5,-8,B),
circle(-3,6,.12), locate(-3,6,C),
circle(-1,4,.12), locate(-1,4,H),

green(line(6,3,-5,-8)),
green(line(6,3,-3,6)),
green(line(-3,6,-5,-8)),
graph( 600, 600, -10, 10, -10, 10,-x+3,-x+3,3x+7)) }}}