Question 1137650

given: {{{DEF}}} with vertices at 
{{{D}}} ({{{-3}}},{{{4}}}) 
{{{E}}} ({{{8}}},{{{-2}}}) 
{{{F}}} ({{{4}}},{{{5}}}) 


the orthocenter: the point where the three "altitudes" of a triangle meet

find  the slopes of  the lines that contain segments {{{DE}}} and{{{ EF}}} 

slope of {{{DE = (-2-4))/(8-(-3)) = -6/11 }}}

now, we need a slope of perpendicular line, altitude to segments {{{DE}}}:

slope of perpendicular line, altitude :  {{{-1/m = -1/(-6/11)}}} -> {{{m = 11/6}}}

{{{y = mx + b}}} (substitute {{{m = 11/6}}}, {{{x = 4}}}, {{{y = 5}}})

{{{5 = (11/6)(4) + b}}}

{{{5 = 44/6 + b}}}

{{{b = 5-22/3}}}

{{{b = 15/3-22/3}}}

{{{b= -7/3}}}


equation of the altitude to {{{DE}}}: 

{{{ y =  (11/6)x -7/3}}}


slope of {{{EF = (-2-4)/(8-5) = -7/4 }}}

altitude to {{{EF}}} : {{{ -1/m = -1/(-7/4)}}} -> {{{m = 4/7}}}

{{{y = mx + b}}} (substitute {{{m = 4/7}}}, {{{x = -3}}}, {{{y =4}}})

{{{4= -3(4/7 ) + b}}}

{{{4 = -12/7 + b}}}

{{{b = 4+12/7}}}

{{{b= 40/7}}}

equation of the altitude to {{{EF}}}:  

{{{y = (4/7) x +40/7}}}
 

intersection of the altitudes will be orthocenter

equation 1:  {{{y =  (11/6)x -7/3}}}
equation 2:   {{{y =(4/7) x +40/7}}}

Solving for {{{x}}} and {{{y}}}:

{{{(11/6)x -7/3=(4/7) x +40/7}}}

{{{(11/6)x -(4/7) x = 7/3+40/7}}}

{{{(53 x)/42= 169/21}}}

{{{x = 338/53}}}

{{{x= 6.4}}}

{{{y =  (11/6)x -7/3}}}

{{{y =  (11/6)(6.4) -7/3}}}

{{{y= 9.4}}}

The coordinates are ({{{6.4}}}, {{{9.4}}}).  This is the orthocenter.

 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-3,4,.12), locate(-3,4,D(-3,4)),
circle(8,-2,.12), locate(8,-2,E(8,-2)),
circle(4,5,.12), locate(4,5,F(4,5)),
line(-3,4,8,-2),line(-3,4,4,5),line(8,-2,4,5),
circle(6.4,9.4,.12), locate(6.4,9.4,O(6.4,9.4)),
 graph( 600, 600, -10, 10, -10, 10, (11/6)(x) -7/3,(4/7) x +40/7)) }}}