Question

Find the orthocenter& circumcenter of a triangle when their vertices are A(1, 2), B(2, 6), C(3,-4).

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Answer 1

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]=
`(6-2)/(2-1)`=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]=
`(-4-2)/(3-1)`=-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=
`(3)/(7)`

again

Substituting value of y in equation 1.

x+4*
`(3)/(7)`+13=0

x=-13-
`(12)/(7)`

x=
`(-103)/(7)`=-14
`(5)/(7)`

So

orthocenter is (-14
`(5)/(7)`,
`(3)/(7)`)

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)=(
`(2+1)/(2)`,
`(2+6)/(2)`)=

(
`(3)/(2)`,4)

Slope of AB=4

Slope of OX=-¼

Equation of line OX passes through (
`(3)/(2)`,4)is

y-4=-¼(x-
`(3)/(2)`)

4y-16=-x+
`(3)/(2)`

8y-16*2=-2x+3

2x+8y=3+32

2x+8y=35

x+4y=
`(35)/(2)`........[1]

again

Y(c,d)=(
`(3+1)/(2)`,
`(-4+2)/(2)`=(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=
`(25)/(2)`

y=
`(25)/(14)`

Substituting value of y in equation 2.

x-3*
`(25)/(14)`=5

x=5+
`(75)/(14)`

x=
`(145)/(14)`

x=10
`(5)/(14)`

circumcenter of a triangle: (10
`(5)/(14)`,1
`(11)/(14)`)