Note:
Ortho centre :a point of intersection of altitudes of a triangle meets the opposite angle.
Given:
For Orthocentre:.
A(1, 2), B(2, 6), C(3,-4). are vertices of a triangle:
Slope of AB[m1]=[tex] \frac{6-2}{2-1} [/tex]=4
Since it is perpendicular to CX.
slope of CX=m2
we have for slope of perpendicular
m1m2=-1
m2=-¼
It passes through the point C(3,-4)
equation of line CX becomes;
(y-y1)=m(x-x1)
y+4=-¼(x-3)
4y+16=-x+3
x+4y+16-3=0
x+4y+13=0........[1]
again:
Slope of AC[m1]=[tex] \frac{-4-2}{3-1} [/tex]=-3
Since it is perpendicular to BY
slope of BY=m2
we have for slope of perpendicular
m1m2=-1
m2=⅓
It passes through the point B(2,6)
equation of line BY becomes;
(y-y1)=m(x-x1)
y-6=⅓(x-2)
3y-18=x-2
x-3y+18-2=0
x-3y+16=0.........[2]
Subtracting equation 1&2.
x+4y+13=0
x-3y+16=0
-__________
7y-3=0
y=[tex] \frac{3}{7} [/tex]
again
Substituting value of y in equation 1.
x+4*[tex] \frac{3}{7} [/tex]+13=0
x=-13-[tex] \frac{12}{7} [/tex]
x=[tex] \frac{-103}{7} [/tex]=-14[tex] \frac{5}{7} [/tex]
So
orthocenter is (-14[tex] \frac{5}{7} [/tex],[tex] \frac{3}{7}[/tex])
And for circumcenter.
Circumcentre: a point of intersection of perpendicular bisector of the triangle.
Now
X,Y and Z are the midpoint of AB,AC and BC respectively.
X(a,b)=([tex] \frac{2+1}{2} [/tex],[tex] \frac{2+6}{2} [/tex])=
([tex] \frac{3}{2} [/tex],4)
Slope of AB=4
Slope of OX=-¼
Equation of line OX passes through ([tex] \frac{3}{2} [/tex],4)is
y-4=-¼(x-[tex] \frac{3}{2} [/tex])
4y-16=-x+[tex] \frac{3}{2} [/tex]
8y-16*2=-2x+3
2x+8y=3+32
2x+8y=35
x+4y=[tex] \frac{35}{2} [/tex]........[1]
again
Y(c,d)=([tex] \frac{3+1}{2} [/tex],[tex] \frac{-4+2}{2} [/tex]=(2,-1)
Slope of AC:-3
Slope of OY=⅓
Equation of line OY passes through (2,-1) is
y+1=⅓(x-2)
3y+3=x-2
x-3y=3+2
x-3y=5......[2]
Multiplying equation 2 by 3 and
Subtracting equation 1&2.
x+4y=35/2
x-3y=5
-_______
7y=[tex] \frac{25}{2} [/tex]
y=[tex] \frac{25}{14} [/tex]
Substituting value of y in equation 2.
x-3*[tex] \frac{25}{14} [/tex]=5
x=5+[tex] \frac{75}{14} [/tex]
x=[tex] \frac{145}{14} [/tex]
x=10[tex] \frac{5}{14} [/tex]
circumcenter of a triangle: (10[tex] \frac{5}{14} [/tex],1[tex] \frac{11}{14} [/tex])
