Question 174559
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Let's draw the triangle with following vertices:
G(-2,5)
H(6,5)
J(4,-1)
{{{drawing(400,400,-3,7,-2,6,graph(400,400,-3,7,-2,6),triangle(-2,5,6,5,4,-1),blue(locate(-2.3,5,G)),blue(locate(6.3,5,H)),blue(locate(4,-1.3,J)))}}}
Now, we draw a line from the vertex that is perpendicular to the opposite side.
1) from point G (-2,5) thru line HJ with points (6,5) & (4,-1):
{{{m[HJ]=(y[2]-y[1])/(x[2]-x[1])=(-1-5)/(4-6)=-6/-2}}}
{{{m[HJ]=3}}}, Slope
Since perpendicular,{{{m[G]=-1/m[HJ]}}}--->{{{m[G]=-1/3}}}
Then,  via Slope-Intercept Form, {{{y=mx+b}}} on point G:
{{{5=(-1/3)(-2)+b}}}--->{{{5=(2/3)+b}}}
{{{b=5-(2/3)=(15-2)/3=13/3}}}
Therefore, the line eqn ----->{{{highlight(y=(-1/3)x+(13/3))}}} , Line passing thru point G perpendicular to Line HJ
2) from point H (6,5) thru line GJ with points (-2,5) & (4,-1):
{{{m[GJ]=(-1-5)/(4-(-2))=-6/(4+2)=-6/6}}}
{{{m[GJ]=-1}}}, Slope
Also, {{{m[H]=-1/m[GJ]}}}---->{{{m[H]=-1/-1=1}}}
Then, via point (6,5):
{{{5=1(6)+b}}}---->{{{b=5-6=-1}}}
Therefore,
{{{highlight(y=1x-1)}}}, Line passing thru point H perpendicular to Line GJ
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3)from point J (4,-1) thru Line GH with points (-2,5) & (6,5):
We can see Line GH has no Slope, only horizontal line (y=5), so line perpendicular to it from point (4,-1) is {{{x+0(y)=4}}},{{{highlight(x=4)}}} vertical line,slope is {{{infinite}}}
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Plug in "highlighted" in the graph:
{{{drawing(400,400,-3,7,-2,6,graph(400,400,-3,7,-2,6,(-1/3)x+(13/3),x-1),triangle(-2,5,6,5,4,-1),blue(locate(-2.3,5,A)),blue(locate(6.3,5,H)),blue(locate(4,-1.3,J)),blue(line(4,-1,4,5)))}}}---------->{{{drawing(400,400,-3,7,-2,6,grid(1),graph(400,400,-3,7,-2,6,(-1/3)x+(13/3),x-1),triangle(-2,5,6,5,4,-1),blue(locate(-2.3,5,A)),blue(locate(6.3,5,H)),blue(locate(4,-1.3,J)),blue(line(4,-1,4,5)),red(circle(4,3,.06)),red(circle(4,3,.08)),red(circle(4,3,.10)),red(circle(4,3,.12)))}}}
Graph on the LEFT, See LINES from one vertex perpendicular to the other side.
Graph on the RIGHT, Lines intersect at one point --->(4,3), orthocenter, ANSWER.
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NEXT
Let's draw the Triangle with following vertices:
Points
A(4,-3)
B(8,5)
C(8,-8)
{{{drawing(400,400,-1,9,-9,6,graph(400,400,-1,9,-9,6),triangle(4,-3,8,5,8,-8),blue(locate(3.7,-3,A)),blue(locate(8.2,5.2,B)),blue(locate(8.2,-8.2,C)))}}}
The same we did the last one,drawing a Line from the vertex that is perpendicular to the opposite side:
1)As you can see on point A (4,-3) is going thru vertical LINE BC and the line from point (4,-3) should be perpendicular to BC---->{{{highlight(y=-3)}}}
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2)Point B (8,5) thru Line AC with points (4,-3) & (8,-8):
{{{m[AC]=(y[2]-y[1])/(x[2]-x[1])=(-8-(-3))/(8-4)=(-8+3)/4}}}}
{{{m[AC]=-5/4}}}
And, {{{m[B]=-1/m[AC]}}}, perpendicular right?
{{{m[B]=-1/(-5/4)=4/5}}}
Then,thru point (8,5): via Slope-Intercept Form {{{y=mx+b}}}
{{{5=(4/5)8+b}}}--->{{{5=32/5+b}}}
{{{b=5-(32/5)=(25-32)/5=-7/5}}}
So, it follows ------->{{{highlight(y=(4/5)x-(7/5))}}}
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3)Point C (8,-8) thru line AB with points (4,-3) & (8,5):
{{{m[AB]=(5-(-3))/(8-4)=(5+3)/4}}}
{{{m[AB]=8/4=2}}}
Then, {{{m[C]=-1/m[AB]=-1/2}}}
Thru point (8,-8):
{{{-8=(-1/cross(2)1)*(cross(8)4)+b}}}----->{{{-8=-4+b}}}
{{{b=-8+4=-4}}}
It follows------>{{{highlight(y=(-1/2)(x)-4)}}}
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Plug in "highlighted" in the graph;
{{{drawing(400,400,-6,9,-9,6,graph(400,400,-6,9,-9,6,-3,(4/5)x-(7/5),(-1/2)x-4),triangle(4,-3,8,5,8,-8),blue(locate(3.7,-3,A)),blue(locate(8.2,5.2,B)),blue(locate(8.2,-8.2,C)))}}}------>{{{drawing(400,400,-6,9,-9,6,grid(1),graph(400,400,-6,9,-9,6,-3,(4/5)x-(7/5),(-1/2)x-4),triangle(4,-3,8,5,8,-8),blue(locate(3.7,-3,A)),blue(locate(8.2,5.2,B)),blue(locate(8.2,-8.2,C)),red(circle(-2,-3,.06)),red(circle(-2,-3,.08)),red(circle(-2,-3,.10)),red(circle(-2,-3,0,0.12)),red(circle(-2,-3,0.16)))}}} 
-------> Orthocenter lies outside the Triangle, being "Obtuse Triangle" >>>> point (-2,-3), ANSWER
Thank you,
Jojo</pre>